Potential/Kinetic Energy Problem

[mhildreth]

Homework Statement

A .2kg bouncy ball loses some of its kinetic energy when it collides with
the ground. Assuming it only maintains 90% of its energy after each bounce, how
high will it get for each of the first three bounces if it is initially launched
vertically at 15 m/s.

Homework Equations

KE = 1/2 mv^2

The Attempt at a Solution

KE(.9) = 1/2(.2)(15)^2

 

[ahmadmz]

The KE you have there is the initial energy. What happens to it after the first bounce? What do you know about Kinetic and Potential energies that you can use to find the first height?

 

[donhubix]

A) Find the first time and height of the ball after the projection upwards:

compare energies to find height:
mgh=mv^2/2 where: m cancelles out, v=15 and g =9.81, find h(height)

Alternativelly:

Find time for the ball to stop from:

V(t)=Vot-at where: a=g, Vo=15, V(t)=0

Having the time you can find the height it will reach:

S=Vot-((at)^2)/2) where: Vo=15, t=time from up(^), a=g.

Having the height use: Ep=mgh to find the potential energy of the ball.

After one bounce the energy will be equal to 90% of the initial mgh value, after two bounces the energy will be mgh*(0.9)^2 and after three mgh*(0.9)^3.

If anything needs to be explained more clearly, just post a reply.