Average force?

Question

Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping 9 m into 30 cm of water. Assuming that he stops just as he reaches the bottom of the water, find the magnitudes of the average force and the total impulse on him from the water. Assume his mass is 60 kg.

I calculated the total impulse to be 783.5 N*s. What would the average force be? I know it's impulse/change in time but I am having trouble calculating it. Thanks for your help!

Answer 1

This sounds like a physics experiment I would rather not try...

Ok, so we use K = U so that 1/2 m v^2 = m g h. Rearrange this and we get

v=sqrt(2 g h) = sqrt (2 * 9 * 9.81) = 13.29 m/s when he hits the water.

His mass is 60 kg, so his initial momentum is 13.29 * 60 = 797.3 kg.m/s. This momentum is lost in the decelleration after he hits the water. (I'd check how you got 783.5 kg m/s)

If we assume constant decelleration (this is a bit of a stretch - but it will work for calculating the average), we can use v^2=u^2-2as. v = 0, u = 13.29 m/s; s = 30 cm and a is what we need.

a = u^2 / 2s = 294.4 ms^-2 - that's about 30g.

F = m a = 60 kg * 294.4 ms^-2 = 17 662N, or to the accuracy of this calculation around 17 700N.

Ouch. That's gotta sting!

hth