# Inelastic collision of ball drop

1. Jan 4, 2006

### nrc_8706

a ball is dropped from rest at the top of a 6.10m-tall building, falls straight downward, collides inelastically with the ground, and bounces back. the ball loses 10.0% of its kinetic energy every time it collides with the ground. how many bounces can the ball make and still reach a window sill that is 2.44m above the ground?

all that i know is that K=1/2mv^2

2. Jan 4, 2006

### mrjeffy321

If the collision is perfectly elastic, the ball will reach its original height after each bounce. but since each bounce, 10% of its energy is lost, (90% remaining), it will only reach 90% of the last height it acheived.

For this question, I dont think it is necesary to actually calculate any kinetic energies or momentums, ...

Can you think of an equation that will relate the rebound height of the ball to the number of bounces that have occured.

3. Jan 5, 2006

### andrevdh

Consider that the potential energy at the top of the bounce is converted to the kinetic energy at the bottom - after the first bounce we have this amount of kinetic energy left then $0.9mgh_o$. This kinetic energy is then converted into potential energy at the top of the bounce $mgh$ for the ball reaching a height $h$ with this available kinetic energy. After the next bounce the total kinetic energy available for the next peak will therefore be $0.9^2mgh_o$. For n bounces it will therefore be $0.9^nmgh_o$. The equation therefore becomes
$$0.9^nh_o=h$$
quite simple is'nt it?

Last edited: Jan 5, 2006