A 63.7 kg person jumps from rest off a 2.90 m-high tower straight down into the

Question

A 63.7 kg person jumps from rest off a 2.90 m-high tower straight down into the water. Neglect air resistance during the descent. She comes to rest 1.00 m under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.

& also another question please:

A 77.8 kg man steps out a window and falls (from rest) 1.04 m to a sidewalk. What is his speed just before his feet strike the pavement?

If the man falls with his knees and ankles locked, the only cushion for his fall is a 0.528 cm give in the pads of his feet. Calculate the average force exerted on him by the ground in this situation. This average force is sufficient to cause cartilage damage in the joints or to break bones.

Explanation / Answer

using work -energy theorem for getting its velocity just before touching water ,
mgh = mu^2 /2
9.81 x 2.90 = u^2 /2
u = 7.54 m/s
now in water ,
using v^2 - u^2 = 2ad
0 - 7.54^2 =2 x a x 1
a = 28.45 m/s
now balancing forces in vertical direction,
Fnet = Fwater -mg = ma
Fwater = m ( a +g) = 63.7 ( 9.81 + 28.45) = 2613.09 N

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