Bouncing Ball

Homework Statement

A little ball bounces straight on a surface in a certain way so that at every bounce it loses a fraction of it's vertical speed (called e in this equation). Assume that the ball is released from a ''resting height' (called H in this equation).

Show that after the n'th bounce against the surface the ball has reached:

$H_n=e^2n * H$

I'm not sure.

The Attempt at a Solution

I'm thinking I need to maybe look at the kinectic energy but I'm sort of mystified. Any help is appreciated.

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Homework Statement

A little ball bounces straight on a surface in a certain way so that at every bounce it loses a fraction of it's vertical speed (called e in this equation). Assume that the ball is released from a ''resting height' (called H in this equation).

Show that after the n'th bounce against the surface the ball has reached:

$H_n=e^2n * H$

I'm not sure.

The Attempt at a Solution

I'm thinking I need to maybe look at the kinectic energy but I'm sort of mystified. Any help is appreciated.
It is supposed to be e^2n not e^2 n

Consider the moment before impact with ground and the moment after impact.

Before impact, speed of the ball = v; Kinetic energy of the ball = 1/2 mv2
After impact, speed of the ball = e*v; Kinetic energy of the ball = 1.2 m (e*v)2

∴the total mechanical energy of the ball has been reduced to e2 of the original after one bounce, and similarly would be reduced to e2+2, after third bounce to e2+2+2, and so on.

∴after the nth bounce mechanical energy of the ball is E=e2n*E0, where E0 is the energy before the first bounce.

As all kinetic energy after impact is then converted into gravitational potential energy when the ball goes to the highest point, and because gravitational potential energy is proportional to the change of height...

Gravitational potential energy is reduced to e2n of the original.
∴the height is Hn =e2n∗H

haruspex