Question

A ball is dropped on the floor from a height of 20m . It rebounds to a height of 10 m . If the ball is in contact with the floor for 0.1 seconds, what is the average acceleration during contact?

Answer 1

a = v2 - v1 / t (1)
consider the case when the ball hits the floor from a height of 20m.
u = 0
v = v1
a = g
s = h1

v^2 = u^2 + 2as
ie, v1^2 = 2gh1
therefore v1 = root(2gh1)

consider the case when the ball bounces to a height of 10m.
u = v1
v = 0
a = -g
s = h2

so, v2 = root(2gh2)

substituting in (1), a = ( root(2gh2) - root(2gh1) ) / t

ie, a = ( root(2 * 9.8 * 10) - root(2 * 9.8 * 20) / 0.1
= 57.98 m/s^2 (upwards)