# Question #2840e

Apr 14, 2017

$2.24 m {s}^{-} 1$, 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
${v}^{2} - {u}^{2} = 2 g h$
If the object is just dropped we have
${v}^{2} = 2 g h$
$v = \sqrt{2 g h}$ ........(1)

Coefficient of restitution $e = \text{velocity of separation"/"velocity of approach}$
Using (1) we get
$e = \frac{\text{velocity after rebound}}{\sqrt{2 g h}}$
$\implies \text{velocity after rebound} = e \sqrt{2 g h}$ ....(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 $= {e}^{2} \sqrt{2 g h}$

Inserting given values and taking $g = 9.81 m {s}^{-} 2$
Rebound velocity after second rebound is $= {\left(0.4\right)}^{2} \sqrt{2 \times 9.81 \times 10}$
$= 2.24 m {s}^{-} 1$, rounded to two decimal places.