physics

You throw a 20-N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is travelling at 25.0 m/s upward. Use the work-energy theorem to find a) its speed just as it left the ground; b) its maximum height.

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  1. knowns:
    F=20N
    h=15m
    V=25m/s

    unknowns:
    max h
    V1

    work-energy theorem: W=K2-K1
    K=.5*mV^2

    so, m(-g)h=.5mV2^2-.5mV1^2

    this simplifies to 2(-g)h=V2^2-V1^2

    therefore, V1=sqrt(V2^2+2gh)

    now all you have to do is plug in numbers.

    V1=sqrt(25^2+2*9.8*15)=30.32m/s

    max height is where V=0

    we can use the work-energy theorem again.

    this time we get: 2(-g)h=0-30.32^2

    so, max h=-(30.32^2)/2(-g)=46.90m

    the final answer is:
    max height = 46.90m
    and initial velocity = 30.32m/s

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  2. Thank you

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