# Help with Work/ Energy Problem

1. Nov 27, 2004

### krypt0nite

Can someone help me explain this problem to me? I try a bunch of things and I dont' really know how to get the answer.
An object starts from rest and slides down a frictionless track as shown. It leaves the track horizontally, striking the ground at a distance d as shown. The same object is now released from twice the height, 2h. How far away will it land?
Diagram of Problem

Somehow... the answer is squareroot 2*d

Last edited: Nov 27, 2004
2. Nov 27, 2004

### prasanna

A very simple problem. No need to worry at all!!!

This problem uses the simple fact that mechanical energy is conserved.
Since it is a frictionless track, the potential energy of the ball on top should
be equal to the kinetic energy when it reaches the bottom of the track.

If m be the mass of the ball and
v be the velocity acquired by the ball at the bottom

m*g*h = 0.5*m*(v^2)

which gives us v = square root(2*g*h)

Now when it leaves the track, its vertical velocity, u, is zero.
Horizontal velocity is v
Let S be the height of the bottom of the track from the ground.
let us calculate the time taken for it to fall from the bottom of the track to the ground.
Remember this time will not depend on the height from which the ball is rolled down.

so

S = (u*t) + 0.5*g*(t^2)
S = 0.5 * g * (t^2)
since u = 0

that gives us

t = square root(2S / g)

now the distance covered horizontally is veloctity*time.
(because there is no acceleration acting in the horizontal direction)

that is d = v*t

d = square root(2gh) * square root(2S / g)

now if the ball is released from 2h, replace the h in the above equation by 2h

that is

new distance = square root(2g*2h) * square root(2S / g)
= squareroot(2) * square root(2gh) * square root(2S / g)

which is nothing but

square root (2) * d