A movie stuntman (mass 78.1 kg) stands on a window ledge 6.19 m above the floor.

Question

A movie stuntman (mass 78.1 kg) stands on a window ledge 6.19 m above the floor. Grabbing a rope attached to a chandelier, he swings down to grapple with the movie’s villain (mass 60.3 kg), who is standing directly under the chandelier. (Assume that the stuntman’s center of mass moves downward 6.19 m. He releases the rope just as he reaches the villain) If the coefficient of kinetic friction of their bodies with the floor is u_k_ = 0.53, how far do they slide?

Give your answer in meters to the second decimal place.

Explanation / Answer

    v =(ms/ms+mv)*v0

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