Question

A ball is dropped from a height of 1.20 m and hits the floor. The ball compresses and then reforms to spring upwards from the floor to a height of 0.86 m. The time from the ball's first contact to when it left contact with the floor was 0.091 s. Calculate the ball's acceleration while it is in contact with the floor.

Answer 1

Answer:

$\large \boxed{\text{98 m \cdot s}^{-2}}$

Explanation:

The formula for the velocity of the ball is

$v = \sqrt{2gh}$

1. Velocity at time of impact

$v = -\sqrt{2 \times 9.807 \times 1.20} = -\sqrt{23.54} = -\textbf{4.85 m/s}$

2. Velocity on rebound

The ball has enough upward velocity to reach a height of 0.86 m.

$v = \sqrt{2 \times 9.807 \times 0.86} = \sqrt{16.87} =\textbf{4.11 m/s}$

3. Acceleration

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{4.11 - (-4.85)}{ 0.091} = \dfrac{8.96 }{0.091} =\textbf{98 m \cdot s}^{\mathbf{-2}}\\\\\text{The acceleration while the ball is in contact with the floor is \large \boxed{\textbf{98 m \cdot s}^{\mathbf{-2}}} }$