# Tennis Ball Dropped onto a Hard Floor| Falling Objects| Kinematics| College Physics| Problem 2.58

### A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (0.0035 s). (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

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

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

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

$\Delta y=0.0051\:m$