# Homework Help: Dropping a Ball off a cliff collision

1. Aug 30, 2009

### George3

1. The problem statement, all variables and given/known data
A ball is dropped from the top of a 48.0 m-high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 26.0 m/s. The stone and ball collide part way up. How far above the base of the cliff does this happen?

2. Relevant equations

3. The attempt at a solution
How would you attempt to start this problem? I know that the stone reaches 34.4 meters as a max height.

2. Aug 30, 2009

### G01

HINT:

You have two unknown variables in this problem, the position of the collision, and the time of the collision. Thus you need to equations to solve the problem.

Set up kinematic equations for the motion of each object and you'll have your two equations. Can you set up these equations?

3. Aug 30, 2009

### George3

This what i did:
For the stone 0 = 26m/s - 9.8m/s^2 t
t= 26/9.8 secs
For the ball: d = (1/2)(9.8m/s^2) * t^2
d = (1/2)(9.8m/s^2) * (26/9.8 secs)^2
d = 34.5m
But that is the wrong answer.

Last edited: Aug 30, 2009
4. Aug 30, 2009

### G01

Your problem is that you have both objects starting from position x=0. This can't be the case. You are working with two different coordinate systems at the same time.

If the stone starts at x=0, then the ball starts at x=48.

5. Aug 30, 2009

### George3

Which kinematic equations would i use?

6. Aug 30, 2009

### rl.bhat

For ball
d1= v1t + 1/2*g*t2

For stone
d2= v2t - 1/2*g*t2

The height of the cliff = d1 + d2