Question

part 3: the force of gravity
forces are not always exerted on objects by direct physical contact, such as a hand push - ing a door closed. for instance, the earth exerts the force of gravity on objects even when the objects are not directly touching the ground. the acceleration on an object due to the force of gravity is 9.8 m/s² downward. in other words, for every second an ob - ject is falling, its velocity increases by 9.8 m/s downward.

1. a. a 9 kg bowling ball rolls off a table and strikes the ground. if the ball is in the air for 0.5 seconds, how fast is the ball moving when it hits the ground?

b. another bowling ball with one - fifth less mass rolls off the same table and strikes the ground. when this ball hits the ground, is it moving faster, slower, or the same speed as the first ball? explain your answer.

Explanation:

Part a

Step1: Identify the formula

We use the kinematic equation \(v = v_0+at\). For an object in free - fall near the Earth's surface, the initial vertical velocity \(v_0 = 0\ m/s\) (assuming it rolls off the table horizontally), and the acceleration \(a = g=9.8\ m/s^{2}\), and the time \(t = 0.5\ s\).

Step2: Substitute the values into the formula

Substitute \(v_0 = 0\ m/s\), \(a = 9.8\ m/s^{2}\), and \(t = 0.5\ s\) into \(v=v_0 + at\).

$$v=0+9.8\times0.5$$

$$v = 4.9\ m/s$$

The acceleration due to gravity \(g = 9.8\ m/s^{2}\) is independent of the mass of the object. Using the kinematic equation \(v=v_0+at\) (with \(v_0 = 0\ m/s\) for horizontal - roll - off and \(a = g\)), the final velocity \(v\) only depends on \(g\) and \(t\). Since both balls are in free - fall (roll off the same table, so same height and thus same time \(t\) in the air, as \(h=\frac{1}{2}gt^{2}\) and \(h\) is the same for both balls) and \(g\) is constant near the Earth's surface.

Answer:

The ball is moving at a speed of \(4.9\ m/s\) when it hits the ground.

Part b