# College Physics 2.58 – A tennis ball dropped onto a hard floor

## Solution:

### Part A

This problem is the same as Problem 2.56. The velocity of the ball when it strikes the ground is

$v=\sqrt{\left(v_o\right)^2+2a\Delta y}$

$v=\sqrt{\left(0\:m/s\right)^2+2\left(-9.80\:m/s^2\right)\left(-1.50\:m\right)}$

$v=5.42\:m/s\:\left(downward\right)$

### Part B

The velocity of the ball just after it leaves the floor on its way back up

$v_o=\sqrt{v^2-2a\Delta y}$

$v_o=\sqrt{\left(0\:m/s\right)^2-2\left(-9.80\:m/s^2\right)\left(1.10\:m\right)}$

$v_o=4.64\:m/s$ upward

### Part C

The acceleration is

$a=\frac{v-v_o}{t}$

$a=\frac{4.643\:m/s-\left(-5.422\:m/s\right)}{0.0035\:s}$

$a=2880\:m/s^2$

### Part D

$\Delta y=\frac{\left(v_o\right)^2-v^2}{2a}$

$\Delta y=\frac{\left(-5.422\:m/s\right)^2-\left(0\:m/s\right)^2}{2\left(2880\:m/s^2\right)}$

$\Delta y=0.0051\:m$