Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 13 m into 35 cm. of water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass to be 72 kg, find the magnitudes of the impulse on him from the water.

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Jayden-James Duffy · Accepted Answer

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Jeffrey Jordon · Accepted Answer

Given that  The height of belly-flop, h=13 m  mass of Henri LaMothe, m=72 kg  The potential energy of him at a height h is PE=mgh  Let the initial velocity when he reaches the water surface =v0  The kinetic energy at the water surface, KE=12mv02  From the law of conservation of energy  12mv02=mgh  v0=2gh  =2(9.8 m/s2)(13 m)  =15.96 m/s  when he reaches the bottom of the water, final velocity vf=0 m/s  Therefore, the change in momentum is given by  △p=m(vf−v0)  =(72 kg)(0−15.96 m/s)  =1149.12 kg⋅m/s  Therefore, the magnitude of the impulse on him from water is  |△p|=|−1149.12 kg⋅m/s|  =1149.12 N⋅s

Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 13 m into 35 cm. of water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass to be 72 kg, find the magnitudes of the impulse on him from the water.

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