Question

A ball hits a wall horizontally at 6.0m/s. It rebounds horizontally at 4.4m/s. The ball is in contact with the wall for 0.040s. What is the acceleration of the ball?

Answer 1

initial velocity = 6 m/s ( forward)
final velocity = 4.4 m/s ( backward) = -4.4 m/s ( forward )

acceleration = change in velocity/time
= ( -4.4 -6)/ 0.04 = -10.4/0.04
= -1040/4 = -260 m/s² ( forward )

Answer 2

The acceleration of the ball is $260 \mathrm{m} / \mathrm{s}^{2}$

To find:

The acceleration of the given ball.

Solution:

From the given,  

We know that,

As the initial velocity and final velocities are opposite in direction,

Initial velocity, u = 6.0 m/s,

Final velocity, v = - 4.4 m/s,  

Time, t = 0.04 s

By using the below formula,  

 Acceleration (a) = $\frac{\text {Final velocity }(v)-\text {Initial velocity }(u)}{{time}(t)}$                                                      

$\mathrm{a}=\frac{-4.4-6.0}{0.04}$

$=\frac{10.4}{0.04}$

=$260\mathrm{m}/\mathrm{s}^{2}$

Acceleration of the given ball  $= 260 \mathrm{m} / \mathrm{s}^{2}$.