Asked by Carden

A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 27 m high. When it hits the ground at the base of the cliff the rock has a speed of 24 m/s.

(a) Assuming that air resistance can be ignored, find the initial speed of the rock.

(b) Find the greatest height of the rock as measured from the base of the cliff.

I know I have to use projectile motion somehow. I don't know how to without knowing the initial velocity. I can't find that without solving the first part. I'm having trouble picking an equation. Please help.
Answered by bobpursley
Initial KE+ initialPE=final KE
I suspect you can solve it from that.

Greatest height? Final KE= greatest PE
solve for greatest height.
Answered by Carden
KE initial = 0
PE initial = mgh = 68.8662 = KE final
KE final = 1/2mv2
v = 23.01

That isn't right. What have I done wrong?
Answered by bobpursley
Where did you get initial KE is zero? That is what you are solving for. You need a tutor, pronto.
Answered by Carden
I am just so confused by this problem. I have done other problems with ease and I'm having so much trouble with other ones.
Answered by Kelly
you do:
1/2mv(initial)^2+mgh=1/2mv(final)^2

so simplify that to : v(inital)= sqrt(v(final)^2-2gh)

For b): 1/2mv(initial)^2=mgh
solve for h : h=v^2/2g
then add your original height to the new height for your answer
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