# College Physics 2.56 – A steel ball drops and rebounds onto a hard floor

## Solution:

### Part A

We are given an initial velocity of 0. We are required to solve for the final velocity. We shall use the formula

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

Then, we substitute the given values

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

$v=5.42\:m/s$

The velocity is 5.42 m/s downward.

### Part B

Consider the floor to be the initial position and the top of the trajectory to be the final position. We shall use the formula

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

The initial velocity is the unknown, so we solve the equation in terms of the initial velocity.

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

Substitute the given values

$v_o=\sqrt{\left(0\:m/s\right)^2-2\left(-9.80\:m/s^2\right)\left(1.45\:m\right)}$

$v_o=5.33\:m/s$

The initial velocity is 5.33 m/s upward.

### Part C

The acceleration is calculated using the formula

$v=v_o+at$

Solve for the equation in terms of acceleration

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

Substitute the given values

$a=\frac{5.331\:m/s-\left(-5.422\:m/s\right)}{8.00\times 10^{-5}\:s}$

$a=1.34\times 10^5\:m/s^2$

### Part D

The period of compression occurs when the ball goes from -5.42 m/s to 5.33 m/s. The acceleration during this time is found in part C. The compression is solve using the formula

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

Solve for Δy in terms of the other variables

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

Then we substitute the given values

$\Delta y=\frac{\left(5.42\:m/s\right)^2-\left(0\:m/s\right)^2}{2\left(1.34\times 10^5\:m/s^2\right)}$

$\Delta y=-1.09\times 10^{-4}\:m$