# If a ball is initially dropped from height h and has coefficient of restitution 0.9 .How many bounced will the ball make before comping to rest? Please help...

Jun 5, 2018

See below

#### Explanation:

If the ball was dropped from a height $h$, then it originally had gravitational potential energy relative to ground, of:

• $U = m g h$.

So, just before it's first collision with ground, the ball will have kinetic energy $T$, such that:

• $T = m g h$

The Coefficient of Restitution, $e$, as it applies between 2 bodies [here, ground and ball], is related to kinetic energy, as:

$e = \sqrt{\left\{\sum {T}_{\text{post-collision"))/{sum T_("pre-collision}}\right\}}$

In the reference frame of ground, only the ball has Kinetic Energy.

$\implies e = \sqrt{\frac{T '}{T}} = 0.9$

$\implies T ' = T \cdot {0.9}^{2}$

$\implies m g h ' = m g h \cdot {0.9}^{2}$

$\implies h ' = 0.81 h$

And so $h '$ is the height to which the ball will rise after the first collision with ground

So:

• ${h}_{1} = 0.81 {h}_{o}$

Follows that:

• ${h}_{2} = 0.81 {h}_{1} = {0.81}^{2} {h}_{o}$

• ${h}_{3} = 0.81 {h}_{2} = {0.81}^{3} {h}_{o}$

• $\ldots$

• ${h}_{n} = {0.81}^{n} {h}_{o}$

In this highly idealised model, the ball will never come to rest, ie ${h}_{n} = 0$ requires that $n \to \infty$.