Question

To test the quality of a tennis ball, you drop it onto a floor from a height of 4.00m . It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0ms,
(a) What is the magnitude of its average acceleration during the contact?
(b) is the acceleration up or down?

Answer 1

Answer:

Part a)

$a = 1210.3 m/s^2$

Part b)

Since it rebounds upwards so it's acceleration is in upwards direction

Explanation:

Velocity of the ball just before it will collide with the floor when it is dropped from height 4.00 m

So we will have

$v_f^2 - v_i^2 = 2 a d$

so we have

$v_1^2 - 0 = 2(9.81)(4.00)$

$v_1 = 8.86 m$

When ball rebound to height h = 2.00 m

so here we have

$v_f^2 - v_i^2 = 2a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

For magnitude of average acceleration

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12\times 10^{-3}}$

$a = 1210.3 m/s^2$

Part b)

Since it rebounds upwards so it's acceleration is in upwards direction

Answer 2

In this problem, we apply the equation regarding  kinematics  expressed as  vf^2 = v0^2 + 2as  vf eventually becomes zero because the ball stops in the end. a = -9.8 m/s2s = 2 metres this time 
This gives initial velocity, vo equal to 6.26m/s 
now 6.26-(-8.85) = 15.11m/s 

change in velocity/change in time = average acceleration 15.11/(12/1000) = 1259.167 m/s^2