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

A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 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 0.0800 ms.

(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?


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



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



The initial velocity is 5.33 m/s upward. 

Part C

The acceleration is calculated using the formula


Solve for the equation in terms of acceleration


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