Until he was in his seventies Henri LaMothe (mass = 60 kg) excited audiences by

Question

Until he was in his seventies Henri LaMothe (mass = 60 kg) excited audiences by belly-flopping from a height of 40 ft into a pool of water 0.3 m deep. Assuming he stopped over the depth of the pool, calculate the following. His momentum just before impact with the water The change in momentum he underwent during impact with the water The impulse he experienced during impact with the water The stopping time of the impact The (average) force on his body during impact How many more times than his weight was the impact with the water? (These are the g forces that he experienced.) Momentum Estimation problems: Choose ONE problem from problems #1-5 and solve it. Be sure to justify the estimations that you had to make to solve the problem.

Explanation / Answer

1ft = 0.3048 m

40 ft = 40*0.3048 = 12.192 m

speed of the person before impact with water = v = sqrt(2*g*h) = sqrt(2*9.81*12.192) = 15.46 m/s

A) momentum just before impact with the water = m*v = 60*15.466 = 928 kg*m/s

B) change in momentum = pf-pi

since he was started form rest initial momentumn pi = 0 m/s

then change in momentum is = pf-o = 928 kg*m/s

C) impulse = change in momentum = 928 N-s

D) before impact speed is v1 = 15.46 m/s

aftr travelling 0.3 m,the speed is v2 = 0 m/s

distance travelled is S = 0.3 m

accelaration a = (v2-v1)/(2*S) = -15.46/(2*0.3) = -25.76 m/s^2

then using time t = (v2-v1)/a = (15.46)/25.76 = 0.6 S

e) Favg*t = impulse = 928

Favg = 928/0.6 = 1546.67 N

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