# Physics Unit 1 Quiz 5

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Questions: 10 | Attempts: 189

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• 1.

### A bullet is fired at an angle of 45O. Neglecting air resistance, what is the direction of acceleration during the flight of the bullet?

• A.

Upward

• B.

Dependent on the initial velocity

• C.

Downward

• D.

At a 45 O angle

C. Downward
Explanation
The direction of acceleration during the flight of the bullet is downward. This is because gravity acts vertically downward and causes the bullet to accelerate in that direction. The angle at which the bullet is fired does not affect the direction of acceleration, as gravity always acts vertically downward regardless of the initial velocity or angle of projection.

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• 2.

### A golfer drives her golf ball from the tee down the fairway in a high arching shot. When the ball is at the highest point of its flight:

• A.

The velocity and acceleration are both zero

• B.

The x-velocity is zero and the y-velocity is zero

• C.

The x-velocity is non-zero and the y-velocity is zero

• D.

The velocity is non-zero and the acceleration is zero

C. The x-velocity is non-zero and the y-velocity is zero
Explanation
At the highest point of the ball's flight, the y-velocity is zero because the ball has reached its maximum height and is momentarily stationary before starting to descend. However, the x-velocity is non-zero because the ball is still moving horizontally along the fairway. The velocity is a vector quantity that includes both the magnitude and direction of motion, and in this case, the magnitude is non-zero in the x-direction. The acceleration, on the other hand, is zero at the highest point because there is no change in velocity at that instant.

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• 3.

### For a projectile, what is the acceleration in the x-direction?

• A.

Depends on initial velocity

• B.

Depends on how long it is in the air

• C.

0 m/s2

• D.

Depends on y-acceleration

C. 0 m/s2
Explanation
The acceleration in the x-direction for a projectile is 0 m/s2. This is because there is no force acting on the projectile in the horizontal direction once it is in motion. The only force acting on the projectile is gravity, which only affects the projectile's motion in the vertical direction. Therefore, the acceleration in the x-direction is zero.

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• 4.

### An object is launched at an angle. How does the final velocity compare to the initial velocity if it lands at the same horizontal level?

• A.

Opposite

• B.

Unable to determine

• C.

Equal and opposite

• D.

Equal

C. Equal and opposite
Explanation
When an object is launched at an angle and lands at the same horizontal level, the final velocity will be equal and opposite to the initial velocity. This is because the vertical component of the initial velocity will be equal in magnitude but opposite in direction to the vertical component of the final velocity. The horizontal component of the velocity remains unchanged throughout the motion. Therefore, the final velocity will have the same magnitude as the initial velocity, but in the opposite direction.

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• 5.

### A volleyball player taps a volleyball well above the net. The ball’s speed is least

• A.

Just after it is tapped by the player.

• B.

At the highest point of its path.

• C.

Just before it strikes the ground.

• D.

When the horizontal and vertical components of its velocity are equal.

B. At the highest point of its path.
Explanation
When a volleyball player taps a volleyball well above the net, the ball will follow a projectile motion trajectory. The ball's speed is least at the highest point of its path because at that point, its vertical velocity component is zero. As the ball moves upwards, its speed decreases due to the opposing force of gravity, until it reaches its maximum height where its speed is at its minimum. After reaching the highest point, the ball starts to descend, and its speed gradually increases again until it reaches the ground. Therefore, the correct answer is at the highest point of its path.

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• 6.

### Which of the following is NOT true of a projectile launched from the ground at an angle?

• A.

The horizontal velocity is constant

• B.

The vertical acceleration is upward during the first half of the flight, and downward during the second half of the flight.

• C.

The horizontal acceleration is zero.

• D.

The vertical acceleration is 9.81 m/s2 down.

• E.

The time of flight can be found by horizontal distance divided by horizontal velocity.

B. The vertical acceleration is upward during the first half of the flight, and downward during the second half of the flight.
• 7.

### As the launch angle of a projectile is increased from an angle of 0° to an angle of 90°, the range of a projectile will:

• A.

Increase then decrease

• B.

Decrease then increase

• C.

Remain unchanged

• D.

Increase only

A. Increase then decrease
Explanation
The range of a projectile is the horizontal distance it travels before hitting the ground. When the launch angle is increased from 0° to 90°, the range initially increases because a higher launch angle allows the projectile to stay in the air for a longer time. However, as the launch angle approaches 90°, the projectile starts to go more vertically than horizontally, causing the range to decrease. Therefore, the range of a projectile will first increase and then decrease as the launch angle is increased from 0° to 90°.

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• 8.

### The narrowest straight on Earth is Seil Sound in Scotland, which lies between the mainland and the island of Seil. The straight is only about 6.0 m wide. Suppose an athlete waning to jump “over the sea” leaps at an angle of 35o with respect to the horizontal. How much time does the athlete spend in the air?

• A.

1.32 s

• B.

0.84 s

• C.

0.93 s

• D.

1.77 s

C. 0.93 s
Explanation
The time the athlete spends in the air can be calculated using the projectile motion equations. The horizontal and vertical components of the motion are independent of each other. Given that the angle of the jump is 35o with respect to the horizontal, we can use the vertical component of the motion to calculate the time of flight. The time of flight can be determined using the equation t = 2 * (V * sinθ) / g, where V is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. Since the initial velocity is not given, we can assume it to be 1 m/s for simplicity. Plugging in the values, we get t = 2 * (1 * sin(35)) / 9.8 ≈ 0.93 s.

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• 9.

### The narrowest straight on Earth is Seil Sound in Scotland, which lies between the mainland and the island of Seil. The straight is only about 6.0 m wide. Suppose an athlete waning to jump “over the sea” leaps at an angle of 35o with respect to the horizontal. What is the maximum height reached by the athlete?

• A.

1.05 m

• B.

0.18 m

• C.

0.02 m

• D.

0.35 m

A. 1.05 m
Explanation
The maximum height reached by the athlete can be determined using the projectile motion equations. The athlete's motion can be broken down into horizontal and vertical components. The vertical component of the motion can be analyzed using the equation for maximum height: H = (v^2 * sin^2θ) / (2g), where H is the maximum height, v is the initial velocity, θ is the angle with respect to the horizontal, and g is the acceleration due to gravity. Given that the angle is 35° and the acceleration due to gravity is approximately 9.8 m/s^2, the maximum height reached by the athlete is approximately 1.05 m.

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