A relative of yours belly flops from a height of 2.50 m (ouch!) and stops moving

Question

A relative of yours belly flops from a height of 2.50 m (ouch!) and stops moving after descending 0.500 m underwater. Her mass is 62.5 kg. (a) What is her speed when she strikes the water? Ignore air resistance. (b) What is the magnitude of her impulse between when she hit the water, and when she stopped? (c) What was the magnitude of her acceleration in the pool? Assume that it is constant. (d) How long was she in the water before she stopped moving? (e) What was the magnitude of the average net force exerted on her after she hit the water until she stopped? (f) Do you think this hurt?

Explanation / Answer

a) v = g.t s = (1/2).g.t² 2.50 = 4.905.t² t = 0.714 v = g.t = 7.00 m/s b) p = m.v = 62.5*7.00 = 438 Nm When she stopped the impulse is 0 (as she is not moving v = 0) c) v = a.t t = 7.00 / a s = (1/2)a.t² 0.500 = (1/2).a.49.00/a² a = 49.0 d) v = a.t 7.00 = 49.00*t t = 0.143 s

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