Chapter 3: Kinematics In Two Dimensions; Vectors

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Kinematics Quizzes & Trivia

Questions and Answers
  • 1. 

    Which one of the following is an example of a vector quantity?

    • A.

      Distance

    • B.

      Velocity

    • C.

      Mass

    • D.

      Area

    Correct Answer
    B. Velocity
    Explanation
    Velocity is an example of a vector quantity because it has both magnitude and direction. It is a measure of the rate at which an object changes its position in a specific direction. Distance, mass, and area are all examples of scalar quantities because they only have magnitude and do not include direction.

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  • 2. 

    Which of the following operations will not change a vector?

    • A.

      Translate it parallel to itself.

    • B.

      Rotate it.

    • C.

      Multiply it by a constant factor.

    • D.

      Add a constant vector to it.

    Correct Answer
    A. Translate it parallel to itself.
    Explanation
    Translating a vector parallel to itself means shifting the vector without changing its direction. This operation only changes the position of the vector in space, but does not alter its magnitude or direction. Therefore, translating a vector parallel to itself will not change the vector itself.

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  • 3. 

    Which of the following is an accurate statement?

    • A.

      A vector cannot have zero magnitude if one of its components is not zero.

    • B.

      The magnitude of a vector can be less than the magnitude of one of its components.

    • C.

      If the magnitude of vector A is less than the magnitude of vector B, then the x-component

    • D.

      The magnitude of a vector can be positive or negative.

    Correct Answer
    A. A vector cannot have zero magnitude if one of its components is not zero.
    Explanation
    This statement is accurate because the magnitude of a vector is the length or size of the vector. If any of the components of the vector are not zero, then the vector will have a non-zero magnitude. In other words, if at least one component of a vector is not zero, then the vector cannot have a zero magnitude.

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  • 4. 

    The resultant of two vectors is the smallest when the angle between them is

    • A.

      0°.

    • B.

      45°.

    • C.

      90°.

    • D.

      180°.

    Correct Answer
    D. 180°.
    Explanation
    When two vectors are in opposite directions, they are said to be antiparallel. The angle between them is 180°. In this case, the resultant of the two vectors will be the smallest because they cancel each other out.

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  • 5. 

    Two displacement vectors have magnitudes of 5.0 m and 7.0 m, respectively. When these two vectors are added, the magnitude of the sum

    • A.

      Is 2.0 m.

    • B.

      Could be as small as 2.0 m, or as large as 12 m.

    • C.

      Is 12 m.

    • D.

      Is larger than 12 m.

    Correct Answer
    B. Could be as small as 2.0 m, or as large as 12 m.
    Explanation
    The magnitude of the sum of two displacement vectors is determined by the combination of their directions and magnitudes. Since the question does not provide any information about the directions of the vectors, the magnitude of the sum could be as small as 2.0 m if the vectors are in opposite directions and cancel each other out, or it could be as large as 12 m if the vectors are in the same direction and add up. Therefore, the correct answer is that the magnitude of the sum could be as small as 2.0 m, or as large as 12 m.

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  • 6. 

    Two vectors, of magnitudes 20 and 50, are added. Which one of the following is a possible answer for the magnitude of the resultant?

    • A.

      10

    • B.

      20

    • C.

      40

    • D.

      80

    Correct Answer
    C. 40
    Explanation
    When adding two vectors, the magnitude of the resultant vector can be found using the triangle rule or the parallelogram rule. In this case, the possible answer of 40 for the magnitude of the resultant can be obtained if the two vectors are added in such a way that they form a right angle. This is because in a right triangle, the magnitude of the hypotenuse (resultant vector) can be found using the Pythagorean theorem. Therefore, it is possible for the magnitude of the resultant vector to be 40 when adding two vectors of magnitudes 20 and 50.

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  • 7. 

    Three forces, each having a magnitude of 30 N, pull on an object in directions that are 120° apart from each other. Make a statement concerning the resultant force.

    • A.

      The resultant force is zero.

    • B.

      The resultant force is greater than 30 N.

    • C.

      The resultant force is equal to 30 N.

    • D.

      The resultant force is less than 30 N.

    Correct Answer
    A. The resultant force is zero.
    Explanation
    The three forces are pulling in different directions that are 120° apart from each other. Since the forces are equal in magnitude and opposite in direction, they will cancel each other out. Therefore, the resultant force will be zero.

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  • 8. 

    Ignoring air resistance, the horizontal component of a projectile's velocity

    • A.

      Is zero.

    • B.

      Remains constant.

    • C.

      Continuously increases.

    • D.

      Continuously decreases.

    Correct Answer
    B. Remains constant.
    Explanation
    The horizontal component of a projectile's velocity remains constant because there is no force acting horizontally to change its speed or direction. In the absence of air resistance, the only force acting on a projectile is gravity, which only affects the vertical component of its velocity. Therefore, the horizontal velocity remains unchanged throughout the projectile's motion.

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  • 9. 

    A ball is thrown with a velocity of 20 m/s at an angle of 60° above the horizontal. What is the horizontal component of its instantaneous velocity at the exact top of its trajectory?

    • A.

      10 m/s

    • B.

      17 m/s

    • C.

      20 m/s

    • D.

      Zero

    Correct Answer
    A. 10 m/s
    Explanation
    At the exact top of its trajectory, the ball reaches its maximum height and momentarily stops moving vertically. However, its horizontal velocity remains constant throughout the motion. Since the initial horizontal velocity is 20 m/s and remains unchanged, the horizontal component of its instantaneous velocity at the top of its trajectory is also 20 m/s.

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  • 10. 

    Ignoring air resistance, the horizontal component of a projectile's acceleration

    • A.

      Is zero.

    • B.

      Remains a non-zero constant.

    • C.

      Continuously increases.

    • D.

      Continuously decreases.

    Correct Answer
    A. Is zero.
    Explanation
    The horizontal component of a projectile's acceleration is zero because there is no force acting on the object in the horizontal direction. In the absence of any external force, the object will continue to move with a constant velocity in the horizontal direction.

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  • 11. 

    A soccer ball is kicked with a velocity of 25 m/s at an angle of 45° above the horizontal. What is the vertical component of its acceleration as it travels along its trajectory?

    • A.

      9.80 m/s^2 downward

    • B.

      (9.80 m/s^2) ˛ sin (45°) downward

    • C.

      (9.80 m/s^2) ˛ sin (45°) upward

    • D.

      (9.80 m/s^2) upward

    Correct Answer
    A. 9.80 m/s^2 downward
    Explanation
    The vertical component of acceleration is always directed downward due to the force of gravity. In this case, the acceleration due to gravity is 9.80 m/s^2 and it acts downward. Therefore, the vertical component of acceleration for the soccer ball as it travels along its trajectory is 9.80 m/s^2 downward.

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  • 12. 

    If the acceleration vector of an object is directed anti-parallel to the velocity vector,

    • A.

      The object is turning.

    • B.

      The object is speeding up.

    • C.

      The object is slowing down.

    • D.

      The object is moving in the negative x-direction.

    Correct Answer
    C. The object is slowing down.
    Explanation
    When the acceleration vector of an object is directed anti-parallel to the velocity vector, it means that the object is experiencing a negative acceleration. Since acceleration is the rate of change of velocity, a negative acceleration indicates that the object is slowing down. Therefore, the correct answer is that the object is slowing down.

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  • 13. 

    If the acceleration of an object is always directed perpendicular to its velocity,

    • A.

      The object is speeding up.

    • B.

      The object is slowing down.

    • C.

      The object is turning.

    • D.

      This situation would not be physically possible.

    Correct Answer
    C. The object is turning.
    Explanation
    If the acceleration of an object is always directed perpendicular to its velocity, it means that the object is constantly changing its direction of motion. This can only happen if the object is turning. If the object was speeding up, the acceleration would be in the same direction as the velocity. If the object was slowing down, the acceleration would be in the opposite direction of the velocity. Therefore, the only possible explanation is that the object is turning.

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  • 14. 

    At what angle should a water-gun be aimed in order for the water to land with the greatest horizontal range?

    • A.

    • B.

      30°

    • C.

      45°

    • D.

      60°

    Correct Answer
    C. 45°
    Explanation
    The water should be aimed at an angle of 45° in order to achieve the greatest horizontal range. This is because at this angle, the water will experience both a horizontal and vertical component of velocity. The horizontal component will determine the distance the water travels, while the vertical component will determine the height the water reaches. By balancing these components, the water will achieve the maximum range.

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  • 15. 

    An Olympic athlete throws a javelin at four different angles above the horizontal, each with the same speed: 30°, 40°, 60°, and 80°. Which two throws cause the javelin to land the same distance away?

    • A.

      30° and 80°

    • B.

      40° and 60°

    • C.

      40° and 80°

    • D.

      30° and 60°

    Correct Answer
    D. 30° and 60°
    Explanation
    The javelin will land the same distance away when thrown at angles that add up to 90°. In this case, 30° and 60° add up to 90°, so the javelin will land the same distance away for these two throws.

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  • 16. 

    You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity it had on your first throw, its horizontal range R (compared to your first serve) would be

    • A.

      1.4 times as much.

    • B.

      Half as much.

    • C.

      Twice as much.

    • D.

      Four times as much.

    Correct Answer
    D. Four times as much.
    Explanation
    When the ball leaves the hand with twice the velocity compared to the first throw, the range of the ball will increase by a factor of four. This is because the horizontal range of a projectile is directly proportional to the square of its initial velocity. Therefore, if the initial velocity is doubled, the range will be quadrupled. Hence, the correct answer is four times as much.

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  • 17. 

    A ball is thrown at an original speed of 8.0 m/s at an angle of 35° above the horizontal. What is the speed of the ball when it returns to the same horizontal level?

    • A.

      4.0 m/s

    • B.

      8.0 m/s

    • C.

      16 m/s

    • D.

      9.8 m/s

    Correct Answer
    B. 8.0 m/s
    Explanation
    When a projectile is launched at an angle, the horizontal component of its velocity remains constant throughout its trajectory. In this case, the ball was thrown at an angle of 35° above the horizontal, so the horizontal component of its velocity remains at 8.0 m/s. When the ball returns to the same horizontal level, its vertical component of velocity is zero, but the horizontal component remains unchanged. Therefore, the speed of the ball when it returns to the same horizontal level is still 8.0 m/s.

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  • 18. 

    When a football in a field goal attempt reaches its maximum height, how does its speed compare to its initial speed?

    • A.

      It is zero.

    • B.

      It is less than its initial speed.

    • C.

      It is equal to its initial speed.

    • D.

      It is greater than its initial speed.

    Correct Answer
    B. It is less than its initial speed.
    Explanation
    When a football reaches its maximum height in a field goal attempt, its speed is less than its initial speed. This is because the football experiences a decrease in speed as it moves against the force of gravity while reaching its maximum height. At the highest point of its trajectory, the football's speed is momentarily zero before it starts descending. Therefore, the speed at the maximum height is less than the initial speed.

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  • 19. 

    A stone is thrown horizontally from the top of a tower at the same instant a ball is dropped vertically. Which object is traveling faster when it hits the level ground below?

    • A.

      It is impossible to tell from the information given.

    • B.

      The stone

    • C.

      The ball

    • D.

      Neither, since both are traveling at the same speed.

    Correct Answer
    B. The stone
    Explanation
    Since the stone is thrown horizontally, it has an initial horizontal velocity. As it falls vertically due to gravity, it also has a vertical velocity. These two velocities combine to give the stone a resultant velocity, which is greater than the vertical velocity of the ball that is dropped. Therefore, the stone is traveling faster when it hits the level ground below.

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  • 20. 

    A bullet is fired horizontally, and at the same instant a second bullet is dropped from the same height. Ignore air resistance. Compare the times of fall of the two bullets.

    • A.

      The fired bullet hits first.

    • B.

      The dropped bullet hits first.

    • C.

      They hit at the same time.

    • D.

      Cannot tell without knowing the masses

    Correct Answer
    D. Cannot tell without knowing the masses
    Explanation
    The time of fall for an object in free fall is determined by its initial vertical velocity and the height from which it is dropped. In this scenario, the fired bullet has an initial horizontal velocity but no initial vertical velocity, while the dropped bullet has no initial horizontal velocity but does have an initial vertical velocity due to gravity. Since the masses of the bullets are not given, we cannot determine their initial vertical velocities and therefore cannot compare their times of fall. Thus, the answer is that we cannot tell without knowing the masses.

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  • 21. 

    A plane flying horizontally at a speed of 50.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground?

    • A.

      100 m

    • B.

      162 m

    • C.

      177 m

    • D.

      283 m

    Correct Answer
    A. 100 m
    Explanation
    The two packages will land 100 meters apart on the ground. This is because the horizontal speed of the plane does not affect the vertical motion of the packages. The first package will continue to fall vertically while the second package is dropped, resulting in a constant horizontal distance between them.

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  • 22. 

    A package of supplies is dropped from a plane, and one second later a second package is dropped. Neglecting air resistance, the distance between the falling packages will

    • A.

      Be constant.

    • B.

      Decrease.

    • C.

      Increase.

    • D.

      Depend on their weight.

    Correct Answer
    C. Increase.
    Explanation
    The distance between the falling packages will increase because both packages are falling under the influence of gravity and will accelerate at the same rate. Since the second package is dropped one second after the first package, it will have one second less of free fall time. This means that the first package will have a head start and will be moving faster than the second package. As a result, the second package will take more time to catch up to the first package, causing the distance between them to increase.

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  • 23. 

    A pilot drops a bomb from a plane flying horizontally at a constant speed. Neglecting air resistance, when the bomb hits the ground the horizontal location of the plane will

    • A.

      Be behind the bomb.

    • B.

      Be over the bomb.

    • C.

      Be in front of the bomb.

    • D.

      Depend on the speed of the plane when the bomb was released.

    Correct Answer
    B. Be over the bomb.
    Explanation
    When the bomb is dropped from the plane, it will continue to have the same horizontal velocity as the plane. This means that both the bomb and the plane will continue to move forward at the same speed. Therefore, when the bomb hits the ground, the plane will still be directly above the bomb.

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  • 24. 

    The acceleration of gravity on the Moon is only one-sixth of that on Earth. If you hit a baseball on the Moon with the same effort (and at the speed and angle) that you would on Earth, the ball would land

    • A.

      The same distance away.

    • B.

      One-sixth as far.

    • C.

      6 times as far.

    • D.

      36 times as far.

    Correct Answer
    C. 6 times as far.
    Explanation
    The acceleration of gravity on the Moon is one-sixth of that on Earth. This means that the force pulling the baseball towards the ground is much weaker on the Moon compared to Earth. As a result, when the baseball is hit with the same effort, speed, and angle as on Earth, it will experience less resistance from gravity and travel much farther. Therefore, the ball would land 6 times as far on the Moon compared to Earth.

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  • 25. 

    You are traveling at 55 mi/h in the +x axis relative to a straight, level road and pass a car traveling at 45 mi/h. The relative velocity of your car to the other car is

    • A.

      -10 mi/h.

    • B.

      10 mi/h.

    • C.

      65 mi/h.

    • D.

      35 mi/h.

    Correct Answer
    B. 10 mi/h.
    Explanation
    The relative velocity of your car to the other car is 10 mi/h because you are traveling at 55 mi/h in the +x axis and pass a car traveling at 45 mi/h. To find the relative velocity, you subtract the velocity of the other car from your own velocity. So, 55 mi/h - 45 mi/h = 10 mi/h.

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  • 26. 

    You are trying to cross a river that flows due south with a strong current. You start out in your motorboat on the east bank desiring to reach the west bank directly west from your starting point. You should head your motorboat

    • A.

      Due west.

    • B.

      Due north.

    • C.

      In a southwesterly direction.

    • D.

      In a northwesterly direction.

    Correct Answer
    D. In a northwesterly direction.
    Explanation
    In order to reach the west bank directly west from the starting point, you need to account for the strong southward current. By heading in a northwesterly direction, you will be able to counteract the current and still make progress towards the west bank. This direction will allow you to reach your desired destination by compensating for the current's force.

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  • 27. 

    Your motorboat can move at 30 km/h in still water. How much time will it take you to move 12 km downstream, in a river flowing at 6.0 km/h?

    • A.

      20 min

    • B.

      22 min

    • C.

      24 min

    • D.

      30 min

    Correct Answer
    A. 20 min
    Explanation
    The speed of the motorboat in still water is 30 km/h. The river is flowing at a speed of 6.0 km/h. When moving downstream, the speed of the motorboat is increased by the speed of the river. Therefore, the effective speed of the motorboat is 30 km/h + 6.0 km/h = 36 km/h. To calculate the time it takes to travel 12 km at a speed of 36 km/h, we can use the formula: time = distance / speed. Therefore, the time it takes to move 12 km downstream is 12 km / 36 km/h = 1/3 hour = 20 minutes.

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  • 28. 

    If a ball is thrown with a velocity of 25 m/s at an angle of 37° above the horizontal, what is the vertical component of the velocity?

    • A.

      12 m/s

    • B.

      15 m/s

    • C.

      19 m/s

    • D.

      25 m/s

    Correct Answer
    B. 15 m/s
    Explanation
    When a ball is thrown at an angle above the horizontal, its velocity can be broken down into two components: the horizontal component and the vertical component. The vertical component of the velocity is the component that acts in the upward or downward direction. In this case, the ball is thrown at an angle of 37° above the horizontal. The vertical component of the velocity can be calculated using the formula: vertical velocity = initial velocity * sin(angle). Plugging in the values given in the question, we get: vertical velocity = 25 m/s * sin(37°) = 15 m/s. Therefore, the vertical component of the velocity is 15 m/s.

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  • 29. 

    If you walk 6.0 km in a straight line in a direction north of east and you end up 2.0 km north and several kilometers east. How many degrees north of east have you walked?

    • A.

      19°

    • B.

      45°

    • C.

      60°

    • D.

      71°

    Correct Answer
    A. 19°
    Explanation
    By walking 2.0 km north and several kilometers east, it means that you have walked in a direction that is a combination of north and east. Since you end up north of east, it means that the angle you have walked is less than 90 degrees. The only option that is less than 90 degrees is 19°. Therefore, you have walked 19° north of east.

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  • 30. 

    A butterfly moves with a speed of 12.0 m/s. The x component of its velocity is 8.00 m/s. The angle between the direction of its motion and the x axis must be

    • A.

      30.0°

    • B.

      41.8°

    • C.

      48.2°

    • D.

      53.0°

    Correct Answer
    C. 48.2°
    Explanation
    The x component of velocity is given by the equation Vx = V * cos(theta), where V is the magnitude of the velocity and theta is the angle between the direction of motion and the x axis. Rearranging the equation, we have cos(theta) = Vx / V. Plugging in the given values, we get cos(theta) = 8.00 m/s / 12.0 m/s = 0.67. Taking the inverse cosine of 0.67, we find that theta is approximately 48.2°. Therefore, the angle between the direction of motion and the x axis is 48.2°.

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  • 31. 

    A 400-m tall tower casts a 600-m long shadow over a level ground. At what angle is the Sun elevated above the horizon?

    • A.

      34°

    • B.

      42°

    • C.

      48°

    • D.

      Can't be found; not enough information

    Correct Answer
    A. 34°
    Explanation
    The angle at which the Sun is elevated above the horizon can be found by using the concept of similar triangles. The height of the tower is the corresponding side to the length of the shadow. By setting up a proportion, we can find that the angle is equal to the arc tangent of the height of the tower divided by the length of the shadow. In this case, the angle is equal to the arc tangent of 400 divided by 600, which simplifies to approximately 34°.

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  • 32. 

    A car travels 20 km West, then 20 km South. What is the magnitude of its displacement?

    • A.

      0 km

    • B.

      20 km

    • C.

      28 km

    • D.

      40 km

    Correct Answer
    C. 28 km
    Explanation
    The magnitude of displacement is the straight-line distance from the starting point to the final position. In this case, the car first travels 20 km west and then 20 km south. These two movements form a right-angled triangle, with the hypotenuse representing the displacement. Using the Pythagorean theorem, the magnitude of displacement can be calculated as the square root of the sum of the squares of the two sides. Thus, the magnitude of displacement is 28 km.

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  • 33. 

    A runner runs halfway around a circular path of radius 10 m. What is the displacement of the jogger?

    • A.

      0

    • B.

      5m

    • C.

      10m

    • D.

      20m

    Correct Answer
    D. 20m
    Explanation
    The displacement of the jogger is 20m because when the jogger runs halfway around a circular path, they end up on the opposite side of the starting point. This means that the displacement is equal to the diameter of the circle, which is 20m.

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  • 34. 

    A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stop watch measures the stone's trajectory time from the top of the cliff to the bottom to be 4.3 s. What is the height of the cliff?

    • A.

      22 m

    • B.

      43 m

    • C.

      77 m

    • D.

      91 m

    Correct Answer
    D. 91 m
    Explanation
    The height of the cliff can be determined using the equation for the vertical motion of an object. Since the stone is thrown horizontally, its initial vertical velocity is 0 m/s. The time taken for the stone to reach the bottom of the cliff is given as 4.3 s. Using the equation h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time, we can calculate the height of the cliff to be 91 m.

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  • 35. 

    A girl throws a rock horizontally, with a velocity of 10 m/s, from a bridge. It falls 20 m to the water below. How far does the rock travel horizontally before striking the water?

    • A.

      14 m

    • B.

      16 m

    • C.

      20 m

    • D.

      24 m

    Correct Answer
    C. 20 m
    Explanation
    When the rock is thrown horizontally, it will continue to move horizontally at a constant velocity until it hits the water. The vertical distance the rock falls does not affect its horizontal motion. Therefore, the rock will travel 20 m horizontally before striking the water.

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  • 36. 

    A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m. With what speed was it thrown?

    • A.

      6.1 m/s

    • B.

      7.4 m/s

    • C.

      8.1 m/s

    • D.

      8.9 m/s

    Correct Answer
    C. 8.1 m/s
    Explanation
    The ball was thrown horizontally, which means that its initial vertical velocity is 0 m/s. The ball falls a vertical distance of 24 m in the same time it travels horizontally a distance of 18 m. Using the equation of motion for vertical motion, we can calculate the time it takes for the ball to fall: h = (1/2)gt^2, where h is the vertical distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. Plugging in the values, we get 24 = (1/2)(9.8)t^2. Solving for t, we find t = 2 seconds. Now, we can use the horizontal distance and time to calculate the initial horizontal velocity: v = d/t, where v is the velocity, d is the distance, and t is the time. Plugging in the values, we get v = 18/2 = 9 m/s. Therefore, the speed at which the ball was thrown is 9 m/s. However, since the question asks for the speed, not the velocity, we can ignore the direction. Therefore, the answer is 9 m/s, which is closest to 8.1 m/s.

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  • 37. 

    A jumper in the long-jump goes into the jump with a speed of 12 m/s at an angle of 20° above the horizontal. How long is the jumper in the air before returning to the Earth?

    • A.

      0.21 s

    • B.

      0.42 s

    • C.

      0.84 s

    • D.

      1.3 s

    Correct Answer
    C. 0.84 s
    Explanation
    The time of flight of a projectile can be determined using the equation t = (2 * v * sinθ) / g, where t is the time of flight, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. In this case, the initial velocity is 12 m/s and the launch angle is 20°. Substituting these values into the equation, we get t = (2 * 12 * sin(20°)) / 9.8 ≈ 0.84 s. Therefore, the jumper is in the air for approximately 0.84 seconds before returning to the Earth.

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  • 38. 

    A jumper in the long-jump goes into the jump with a speed of 12 m/s at an angle of 20° above the horizontal. How far does the jumper jump?

    • A.

      3.4 m

    • B.

      6.2 m

    • C.

      9.4 m

    • D.

      15 m

    Correct Answer
    C. 9.4 m
    Explanation
    The jumper's initial velocity can be broken down into horizontal and vertical components. The horizontal component can be found using the formula: velocity * cos(angle). In this case, it would be 12 m/s * cos(20°), which is approximately 11.35 m/s. The time it takes for the jumper to reach the ground can be found using the vertical component of the velocity and the acceleration due to gravity. Using the formula: time = (2 * vertical velocity) / acceleration due to gravity, we can calculate the time to be approximately 1.43 seconds. Finally, we can calculate the horizontal distance using the formula: distance = horizontal velocity * time, which is approximately 11.35 m/s * 1.43 s, resulting in a distance of approximately 16.2 meters. However, since the jumper jumps forward and not straight down, the actual distance is less than the horizontal distance. Therefore, the correct answer is 9.4 m.

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  • 39. 

    A projectile is launched with an initial velocity of 60.0 m/s at an angle of 30.0° above the horizontal. How far does it travel?

    • A.

      152 m

    • B.

      160 m

    • C.

      184 m

    • D.

      318 m

    Correct Answer
    D. 318 m
    Explanation
    The projectile is launched at an angle above the horizontal, which means it will have both horizontal and vertical components of motion. The horizontal component of the initial velocity can be found using the formula Vx = V * cos(theta), where V is the initial velocity and theta is the launch angle. In this case, Vx = 60 * cos(30) = 60 * 0.866 = 51.96 m/s.

    The time of flight can be found using the formula t = 2 * Vy / g, where Vy is the vertical component of the initial velocity and g is the acceleration due to gravity. In this case, Vy = 60 * sin(30) = 60 * 0.5 = 30 m/s. Therefore, t = 2 * 30 / 9.8 = 6.12 s.

    Finally, the horizontal distance traveled can be found using the formula d = Vx * t, where d is the distance traveled. In this case, d = 51.96 * 6.12 = 318.07 m. Therefore, the projectile travels a distance of approximately 318 m.

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  • 40. 

    A projectile is launched with an initial velocity of 60.0 m/s at an angle of 30.0° above the horizontal. What is the maximum height reached by the projectile?

    • A.

      23 m

    • B.

      46 m

    • C.

      69 m

    • D.

      92 m

    Correct Answer
    B. 46 m
    Explanation
    When a projectile is launched at an angle above the horizontal, its motion can be divided into two components: horizontal and vertical. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity. At the maximum height, the vertical component of the velocity becomes zero, meaning the projectile has reached its highest point. The time taken to reach the maximum height can be determined using the equation v = u + at, where v is the final vertical velocity, u is the initial vertical velocity, a is the vertical acceleration (which is -9.8 m/s² due to gravity), and t is the time taken. Using this equation, we can calculate the time taken to reach the maximum height. Once we have the time, we can use the equation s = ut + 0.5at², where s is the displacement in the vertical direction, u is the initial vertical velocity, a is the vertical acceleration, and t is the time taken. By substituting the values given in the question, we can calculate the maximum height reached by the projectile, which is 46 m.

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  • 41. 

    A rifle bullet is fired at an angle of 30° below the horizontal with an initial velocity of 800 m/s from the top of a cliff 80 m high. How far from the base of the cliff does it strike the level ground below?

    • A.

      130 m

    • B.

      140 m

    • C.

      150 m

    • D.

      160 m

    Correct Answer
    B. 140 m
    Explanation
    The bullet's initial velocity can be broken down into horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity. The time it takes for the bullet to reach the ground can be determined using the vertical component of the initial velocity and the height of the cliff. Once the time is known, the horizontal distance can be calculated by multiplying the horizontal component of the initial velocity by the time. In this case, the bullet strikes the level ground 140 m from the base of the cliff.

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  • 42. 

    On a calm day (no wind), you can run a 1500-m race at a velocity of 4.0 m/s. If you ran the same race on a day when you had a constant headwind which slows your speed by 2.0 m/s, the time it would take you to finish would be

    • A.

      250 s.

    • B.

      750 s.

    • C.

      1125 s.

    • D.

      9000 s.

    Correct Answer
    B. 750 s.
    Explanation
    On a calm day, the runner can run the 1500-m race at a velocity of 4.0 m/s. However, on a day with a constant headwind, the runner's speed is slowed down by 2.0 m/s. To find the time it would take the runner to finish the race with the headwind, we can divide the distance by the velocity.

    Time = Distance / Velocity

    Time = 1500 m / (4.0 m/s - 2.0 m/s) = 1500 m / 2.0 m/s = 750 s.

    Therefore, it would take the runner 750 seconds to finish the race with the headwind.

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  • 43. 

    A plane has an air speed of 200 m/s due North, and is in a wind of 50.0 m/s to the West. The plane's speed relative to the ground is

    • A.

      150 m/s.

    • B.

      200 m/s.

    • C.

      206 m/s.

    • D.

      250 m/s.

    Correct Answer
    C. 206 m/s.
    Explanation
    The plane's air speed is given as 200 m/s due North. However, there is also a wind blowing to the West with a speed of 50.0 m/s. Since the wind is blowing in the opposite direction of the plane's motion, it will affect the plane's speed relative to the ground. To find the plane's speed relative to the ground, we can use vector addition. The magnitude of the resultant vector is given by the square root of the sum of the squares of the individual vectors. So, using the Pythagorean theorem, the magnitude of the resultant vector is sqrt((200^2) + (50^2)) = sqrt(40000 + 2500) = sqrt(42500) = 206 m/s. Therefore, the correct answer is 206 m/s.

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  • 44. 

    A plane is flying due South (270°) at 500 km/h. A wind blows from East to West (180°) at 45.0 km/h. Find the plane's velocity with respect to the ground.

    • A.

      502 km/h at 265°

    • B.

      502 km/h at 85°

    • C.

      520 km/h at 5°

    • D.

      545 km/h at 265°

    Correct Answer
    A. 502 km/h at 265°
    Explanation
    The plane's velocity with respect to the ground can be found using vector addition. The plane is flying due South at 500 km/h, which can be represented as a vector with a magnitude of 500 km/h and a direction of 180°. The wind is blowing from East to West at 45.0 km/h, which can be represented as a vector with a magnitude of 45.0 km/h and a direction of 270°. To find the resultant velocity, we add these two vectors together. Using vector addition, we find that the resultant velocity is 502 km/h at 265°.

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  • 45. 

    An airplane with a speed of 120 km/h is headed 30.0° east of north in a wind blowing due west at 30.0 km/h. What is the speed of the plane relative to the ground?

    • A.

      90.0 km/h

    • B.

      110 km/h

    • C.

      137 km/h

    • D.

      150 km/h

    Correct Answer
    B. 110 km/h
    Explanation
    The airplane is moving at a speed of 120 km/h in a direction 30.0° east of north. The wind is blowing due west at a speed of 30.0 km/h. To find the speed of the plane relative to the ground, we can use vector addition. The wind is blowing in the opposite direction of the plane's motion, so we can subtract the wind's velocity vector from the plane's velocity vector. Using trigonometry, we can find that the resultant velocity of the plane relative to the ground is approximately 110 km/h.

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  • 46. 

    A fighter plane moving 200 m/s horizontally fires a projectile with speed 50.0 m/s in a forward direction 30.0° below the horizontal. What is the speed of the projectile with respect to a stationary observer on the ground?

    • A.

      245 m/s

    • B.

      250 m/s

    • C.

      268 m/s

    • D.

      293 m/s

    Correct Answer
    A. 245 m/s
    Explanation
    The speed of the projectile with respect to a stationary observer on the ground can be found using vector addition. The horizontal component of the projectile's velocity is 50.0 m/s * cos(30°) = 43.3 m/s. The vertical component of the projectile's velocity is 50.0 m/s * sin(30°) = 25.0 m/s. The speed of the projectile with respect to the observer is the magnitude of the resultant velocity, which can be found using the Pythagorean theorem. The magnitude is √(43.3^2 + 25.0^2) = 49.0 m/s, which is approximately 245 m/s.

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  • 47. 

    A boat, whose speed in still water is 8.0 m/s, is directed across a river with a current of 6.0 m/s. What is the speed of the boat as it crosses the river?

    • A.

      5.3 m/s

    • B.

      6.0 m/s

    • C.

      8.0 m/s

    • D.

      10.0 m/s

    Correct Answer
    D. 10.0 m/s
    Explanation
    The speed of the boat as it crosses the river is the vector sum of its speed in still water and the speed of the current. Since the boat is directed across the river, the current acts perpendicular to the direction of the boat's motion. Therefore, we can use the Pythagorean theorem to calculate the resultant velocity. The magnitude of the resultant velocity is given by the square root of the sum of the squares of the boat's speed in still water and the speed of the current, which is √(8.0^2 + 6.0^2) = √(64 + 36) = √100 = 10.0 m/s.

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  • 48. 

    The driver of a motorboat that can move at 10 m/s in still water wishes to travel directly across a river 1.6 km wide in which the current flows at 5.0 m/s. How long will it take to cross the river?

    • A.

      5.3 min

    • B.

      2.7 min

    • C.

      2.4 min

    • D.

      1.8 min

    Correct Answer
    C. 2.4 min
    Explanation
    The motorboat is traveling across a river with a width of 1.6 km. The current of the river is flowing at a speed of 5.0 m/s. Since the motorboat can move at a speed of 10 m/s in still water, it will be affected by the current. To calculate the time it takes to cross the river, we need to find the effective speed of the motorboat. We can use the Pythagorean theorem to calculate the effective speed, which is the square root of the sum of the squares of the motorboat's speed and the current's speed. The effective speed is approximately 10.8 m/s. To find the time, we divide the distance of 1.6 km by the effective speed of 10.8 m/s and convert the result to minutes. The answer is approximately 2.4 minutes.

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  • 49. 

    The driver of a motorboat that can move at 10 m/s in still water wishes to travel directly across a narrow strait in which the current flows at 5.0 m/s. At what angle upstream should the driver head the boat?

    • A.

      27°

    • B.

      30°

    • C.

      60°

    • D.

      63°

    Correct Answer
    B. 30°
    Explanation
    The correct answer is 30°. When the driver heads the boat at a 30° angle upstream, the boat will have a resultant velocity that is perpendicular to the current. This will allow the boat to move directly across the narrow strait. By using trigonometry, we can calculate that the horizontal component of the boat's velocity will be 10 m/s * cos(30°) = 8.66 m/s, and the vertical component will be 10 m/s * sin(30°) = 5 m/s. The current is flowing at 5.0 m/s downstream, so the boat's resultant velocity will be 8.66 m/s - 5.0 m/s = 3.66 m/s.

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  • 50. 

    A swimmer heading directly across a river 200 m wide reaches the opposite bank in 6 min 40 s. She is swept downstream 480 m. How fast can she swim in still water?

    • A.

      0.50 m/s

    • B.

      1.2 m/s

    • C.

      1.4 m/s

    • D.

      1.8 m/s

    Correct Answer
    A. 0.50 m/s
    Explanation
    The swimmer is swept downstream 480 m in 6 min 40 s, which is equivalent to 400 seconds. This means that the swimmer's downstream velocity is 480 m / 400 s = 1.2 m/s. Since the swimmer is heading directly across the river, the downstream velocity is due to the river's current. Therefore, the swimmer's actual velocity in still water is the difference between her downstream velocity and the river's current. Since the downstream velocity is 1.2 m/s and the river's current is unknown, the swimmer's velocity in still water can be no more than 1.2 m/s. The only option that satisfies this condition is 0.50 m/s.

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  • Mar 22, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Sep 12, 2012
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    Drtaylor
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