Question

A ball is initially at rest and it is dropped from a building of height of $10m$ to the ground. If the coefficient of restitution is $e=0.7$ after the collision, the ball rebounds back up to height $h$, find the value of $h$. ($g=9.8m/s^2$)

Answer 1

Step 1: Given Data

Initial velocity $u=0$

Height of the building $S=10m$

Coefficient of restituting $e=0.7$

Rebound height of the blood $=h$

Acceleration due to gravity $g=9.8m/s^2$

Let the velocity of the ball just before reaching the ground be $v_a$.

Let the velocity of the ball after rebound be $v_s$.

Let the final velocity after rebound be $v=0$

Step 2: Calculate $v_a$

From Newton's third equation of motion,

${v_a}^2=u^2+2gS$

$\Rightarrow v_a=\sqrt{2gS}$

$\Rightarrow v_a=\sqrt{20g}$ $\left(1\right)$

Step 3: Calculate $v_s$

From Newton's third equation of motion,

$v^2={v_s}^2+2gh$

$\Rightarrow -v_s=\sqrt{2\left(-g\right)h}$

$\Rightarrow v_s=\sqrt{2gh}$ $\left(2\right)$

Step 4: Calculate $h$

We know that the coefficient of restitution is given as

$e=\frac{v_s}{v_a}=0.7$

$\Rightarrow \frac{\sqrt{2gh}}{\sqrt{20g}}=0.7$

$\Rightarrow \frac{h}{10}=0.49$

$\Rightarrow h=0.49\times 10$

$\Rightarrow h=4.9m$

Hence, the value of $h$ is $4.9m$.