- #1
sunnyday
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Homework Statement
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?
Homework Equations
d = vt+1/2at^2
d1 + d2 = 55
The Attempt at a Solution
Ball[/B]
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.