Physics

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.

asked by Carden
  1. Initial KE+ initialPE=final KE
    I suspect you can solve it from that.

    Greatest height? Final KE= greatest PE
    solve for greatest height.

    posted by bobpursley
  2. 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?

    posted by Carden
  3. Where did you get initial KE is zero? That is what you are solving for. You need a tutor, pronto.

    posted by bobpursley
  4. 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.

    posted by Carden
  5. 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

    posted by Kelly

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