Physics

A 70 kg person is at the top of a 10 m tall slide. Ignoring friction, what is his or her speed at the halfway point where he or she is 5 m from the bottom of the slide? What is his or her speed at the bottom of the slide?

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  1. V^2 = Vo^2 + 2g*d = 0 + 19.6*(10-5)=98
    V = 9.90 m/s.

    V^2 = 0 + 19.6*10 = 196
    V = 14 m/s.

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    posted by Henry

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