Question

A ball is dropped from rest from the top of a 6.10 m-y’all building, falls straight downward, collides inelasrically with the ground, and bounces back. The call 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 windowsill that is 2.44 m above the ground?

Answer 1

E = mgh

You then know that each time it hits the ground, it loses 10.0%, or in other words, it becomes (100 - 10.0 = 90%) the value it was before. If you were to re-calculate the energy after the first bounce (and second and third etc..), you would simply end up multiplying by the same percentage factor.
Because of this, you can just re-write it as:

Ef = (0.90)^n * mgh

where Ef is the final energy, ie. the energy when it is 2.44m off the ground and 'n' is how many bounces it's made, so:

mg(2.44) = (0.90)^n * mg(6.10), cancel the mg's

2.44 = 6.10* (0.90)^n
(.9)^n= .4

n= 8.70
n <= 8.