PHYSICS HELP

A 0.26-kg rock is thrown vertically upward from the top of a cliff that is 31 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s .

a.Assuming that air resistance can be ignored, find the initial speed of the rock .
v0= M/S

b. Find the greatest height of the rock as measured from the base of the cliff.
Hmax= m

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  1. b. V^2 = Vo^2 + 2g*hmax.
    29^2 = 0 + 19.6*hmax,
    hmax = 42.9 m. = Max. ht. above gnd.

    ho + h1 = hmax.
    31 + h1 = 42.9,
    h1 = 11.9 m. above the cliff.

    a. V^2 = Vo^2 + 2g*h1.
    0 = Vo^2 -19.6*11.9,
    Vo^2 = 233.2,
    Vo = 15.3 m/s = Velocity at which the rock was released.

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