SPH3U Midterm Physics Motion Concepts

To find the maximum height reached by the ball, we can use the kinematic equation: \( v^2 = u^2 + 2a s \), where \( v \) is the final velocity (0 m/s at the maximum height), \( u \) is the initial velocity (18 m/s), \( a \) is the acceleration (-9.8 m/s², since it acts downward), and \( s \) is the maximum height. Rearranging the equation gives \( s = \frac{v^2 - u^2}{2a} \). Plugging in the values yields \( s = \frac{0 - (18)^2}{2 \times -9.8} \), resulting in a maximum height of 16.5 m.

To find the displacement of the car, we can use the formula for uniform acceleration: \( s = ut + \frac{1}{2}at^2 \). Here, the initial velocity \( u \) is 0 m/s, the final velocity \( v \) is 25 m/s, and the time \( t \) is 10 seconds. First, we calculate the acceleration \( a \) using \( a = \frac{v - u}{t} = \frac{25 - 0}{10} = 2.5 \, \text{m/s}^2 \). Substituting into the displacement formula gives \( s = 0 + \frac{1}{2} \times 2.5 \times (10^2) = 125 \, \text{m} \).

To find the horizontal range of the projectile, we first calculate the time it takes to fall 45 m using the formula \( h = \frac{1}{2} g t^2 \). Rearranging gives \( t = \sqrt{\frac{2h}{g}} \), which results in approximately 3.02 seconds. The horizontal distance traveled during this time is found by multiplying the horizontal velocity (20 m/s) by the time of flight. Thus, the range is \( 20 \, \text{m/s} \times 3.02 \, \text{s} \approx 60.4 \, \text{m} \).

A straight horizontal line on a velocity-time graph indicates that the velocity of the object remains constant over time. This means the object is not accelerating or decelerating, but rather moving at a uniform speed. Since there are no changes in velocity, the object is in uniform motion, maintaining the same distance covered in equal intervals of time.

On a position-time graph, the slope of the line indicates how quickly the position of an object changes over time. This rate of change is defined as velocity. A steeper slope signifies a higher velocity, while a flatter slope indicates a lower velocity. When analyzing the graph at a specific point, the slope at that point reflects the instantaneous velocity, providing insight into the object's speed and direction at that precise moment.

To find the vertical component of the initial velocity of a projectile, we use the formula: \( V_{y} = V \cdot \sin(\theta) \), where \( V \) is the initial speed and \( \theta \) is the launch angle. In this case, \( V = 30 \, \text{m/s} \) and \( \theta = 40° \). Substituting the values gives \( V_{y} = 30 \cdot \sin(40°) \approx 30 \cdot 0.643 \approx 19.29 \, \text{m/s} \). This calculation shows that the vertical component of the initial velocity is approximately 19.29 m/s.

An object in free fall near Earth's surface is influenced primarily by gravity, which pulls it downward. This gravitational force results in a constant acceleration of approximately 9.8 m/s², regardless of the object's mass. This acceleration remains consistent as long as air resistance is negligible, making it a fundamental principle of classical mechanics. Thus, any object in free fall will accelerate towards the Earth at this rate, confirming the statement as true.