Question #2840e

1 Answer
Apr 14, 2017

Answer:

#2.24ms^-1#, rounded to two decimal places.

Explanation:

A free falling object from a height #h# reaches the ground with a velocity #v# given by the kinematic equation
#v^2-u^2=2gh#
If the object is just dropped we have
#v^2=2gh#
#v=sqrt(2gh)# ........(1)

Coefficient of restitution #e="velocity of separation"/"velocity of approach"#
Using (1) we get
#e="velocity after rebound"/sqrt(2gh)#
#=>"velocity after rebound"=esqrt(2gh)# ....(2)

The object goes up with velocity as in (2) and returns, under gravity, with the same velocity directed downwards.

Therefore, Rebound velocity after second rebound is #=e^2sqrt(2gh)#

Inserting given values and taking #g=9.81ms^-2#
Rebound velocity after second rebound is #=(0.4)^2sqrt(2xx9.81xx10)#
#=2.24ms^-1#, rounded to two decimal places.