Question

Part 1 of the question= A tennis ball is dropped from 1.29 m above the ground. It rebounds to a height of 0.839 m. With what velocity does it hit the ground? The acceleration of gravity is 9.8 m/s ^ 2 . (Let down be negative . ) Answer in units of m/s . Part 2 of the question= With what velocity does it leave the ground? Answer in units of m/s .Part 3 of the question= If the tennis ball were in contact with the ground for 0.0133 s. find the acceleration given to the tennis ball by the ground. Answer in units of m/s^ 2

Answer 1

Answer:

1. 5.03 m/s

2. 4.06 m/s

3. 683.46 m/s

Explanation:

Part 1.

Since the tennis ball is dropped, we can calculate the final velocity using the following equation.

$v_{}=\sqrt[]{2gh}$

Where g is the acceleration of gravity and h is the height. So, replacing the values, we get:

$\begin{gathered} v_{}=\sqrt[]{2(9.8)(1.29)} \\ v=5.03\text{ m/s} \end{gathered}$

Then, the tennis ball hits the ground at 5.03 m/s

Part 2.

Now, we can use the same equation from part 1 but the height this time will be equal to 0.839 m, then, the velocity is equal to:

$\begin{gathered} v=\sqrt[]{2(9.8)(0.839)} \\ v=4.06\text{ m/s} \end{gathered}$

So, the tennis ball leaves the ground at 4.06 m/s

Part 3.

Finally, the acceleration is equal to the change in velocity over time, so:

$a=\frac{4.06-(-5.03)}{0.0133}=683.46m/s^2$

Therefore, the acceleration given to the tennis ball by the ground is 683.46 m/s²