An easy question for some (i hope)

Rnger

"A small rubber ball is held at height h above a smooth level °oor and released from rest
at time t = 0. If the coeffcient of restitution between the ball and the floor is e, show
that after the first bounce the ball rises to a height h1 where h1 = (e^2)h.
The ball continues to bounce until it comes to rest. Show that the total distance travelled by the ball from initial release to rest is (1+e^2)h/(1-e^2) ."

How do I determine the total distance travelled? Thanking you in advance for helping me with my homework

David19

Can you do the first part? What attempt have you made? For the second part I think you end up with a series which you can then find the sum of, thus giving the distance travelled. Can you find the series?

dudara

Part 1: the ball dropping to the ground, through a height h.

u = 0
v = v
a = +g
s = h

v^2 = u^2 + 2*a*s ends up giving v = sqrt(2*g*h)

Part 2: The bounce.

The coefficient means that new speed = e*(old speeed)

therefore the ball leaves the ground with an initial speed of u = e*sqrt(2*g*h)

Part 3: the rise up to the new height h1

u = e*sqrt(2*g*h)
v = 0 (at max height)
a = -g
s = h1

Work it out from there.

As the second part, you have a series with terms of the form

hn = (e^2)*hn-1