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15 May, 18:45

A 240 g ball is dropped from a height of 2.1 m, bounces on a hard floor, and rebounds to a height of 1.6 m. The figure shows the impulse received from the floor. What maximum force does the floor exert on the ball?

impulse =.005*Fmax

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  1. 15 May, 19:48
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    FΔt=mΔv-->F=m (Δv) / (Δt)

    We can find the speed by calculating the potential energy and converting it to kinetic energy. Ep = Ek - - >mgh=m v^2 / 2

    v = (√ 2gh) = (√ 2∗9.81∗2.1) ≈6.42m/s

    so then F=.240∗6.42/0.005=308 N (we know the time because of the above mentioned impulse equation.)
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