A 68 kg diver steps off a 12 m tower and drops from rest straight down intothe w

Question

A 68 kg diver steps off a 12 m tower and drops from rest straight down intothe water. If he comes to rest 4.0 mbeneath the surface, determine the average resistance force exertedon him by the water.
1 N

Explanation / Answer

KEf+PEf=KEi+PEi PEf is 0, because the height will be 0, conversely KEi is 0 becausevelocity is 0. 1/2mvf2=mgy 1/2vf2=gy => divide m out 1/2vf2=9.8m/s2*15m vf=17.14m/s Now using the kinematic formula, we can find the acceleration. vf2=vi2+2ax the velocity above becomes the initial velocity in this equation,and the final velocity becomes 0. 0=(17.14m/s)2+2a(4m) a=-36.72 m/s2 Now, using Newton's second law, F=ma F=(68kg)(-36.72m/s2) F=-2496.96N

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