Question

Question #2840e

Answer 1

`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.