A ball is dropped from the top of a 55.0 m high cliff

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1. Jun 1, 2017

sunnyday

1. The problem statement, all variables and given/known data
A ball is dropped from the top of a 55.0 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 21.0 m/s . The stone and ball collide part way up.

How far above the base of the cliff does this happen?
2. Relevant equations
d = vt+1/2at^2
d1 + d2 = 55

3. The attempt at a solution
Ball

v = 0
d = 1/2*9.8t^2

Stone
v = 21
d = 21t-1/2*9.8t^2 (I made it -9.8 because the stone is going up so it's fighting gravity, I hope I'm right)

Ball + Stone = 55
1/2*9.8t^2 + 21t-1/2*9.8t^2 = 55
21t = 55
t = 2.619

Stone:
21(2.619)-(1/2)(9.8)(2.619)^2 = 21.389

Ball:
(1/2)(9.8)(2.619)^2 = 33.6109

I don't understand what I'm supposed to do next.

2. Jun 1, 2017

PeroK

What do you mean next? You've got the answer, surely?

Note that you need to add units to your working, e.g. $t = 2.619s$.

An an aside: the stone starts off at $21m/s$ and the initial distance between the stone and the ball is $55m$. What do you get if you divide $55$ by $21$?

3. Jun 1, 2017

ehild

Stone:
21(2.619)-(1/2)(9.8)(2.619)^2 = 21.389

Ball:
(1/2)(9.8)(2.619)^2 = 33.6109

I don't understand what I'm supposed to do next.[/QUOTE]
You have to answer the question: How far above the base of the cliff does this happen?
You got two numbers, explain what they mean. First, add units, as @PeroK said.
Clarify, at what distance from the base of the cliff the ball and stone collide.

4. Jun 1, 2017

sunnyday

When I divided 55/21, I got 2.619 (the same number as I got for seconds). I got 2 answers, the displacement for the ball (33.6109m) and the displacement for the stone (21.389m).

5. Jun 1, 2017

PeroK

How far about the cliff is the collision? Hint: think about the stone.

Is $55/21 = 2.619$ a coincidence?

6. Jun 1, 2017

ehild

Answer the question: How far above the base of the cliff does the collision happen?