Dropping a Ball off a cliff collision

[George3]

Homework Statement

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?

Homework Equations

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.

 

[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?

 

[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.

 

[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.

 

[George3]

Which kinematic equations would i use?

 

[rl.bhat]

Which kinematic equations would i use?

For ball
d₁= v₁t + 1/2*g*t²

For stone
d₂= v₂t - 1/2*g*t²

The height of the cliff = d1 + d2