A person jumps from the roof of a house 3.0-m high. When he strikes the ground b

Question

A person jumps from the roof of a house 3.0-m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 m. If the mass of his torso (excluding legs) is 46 kg .

A. Find his velocity just before his feet strike the ground.? Express your answer using two significant figures. Enter positive value if the velocity is upward and negative value if the velocity is downward. I keep getting 7.6 m/s but it is wrong ?

B. Find the average force exerted on his torso by his legs during deceleration.

Express your answer using two significant figures. Enter positive value if the force is upward and negative value if the force is downward.

Explanation / Answer

height h = 3 m
from law of conservation of energy ,energy conservation, we have:

      mgh + 0 = (1/2)mv2+ 0
   velocity v = (2gh)
                  = {(2)(9.8 m/s2)(3 m)}
                  = 7.67 m/s = 7.7 m/s
.....................................................................
displacement S = -0.7 m
from kinematic equations ,
     v2 - v02 = 2aS
    0 - (7.67 m/s)2 = 2aS
    - (7.67 m/s)2 = 2aS
acceleration a = 42 m/s2
net force F = ma
                 = (46 kg)(42 m/s2)
                 = 1932 N (upward)
weight W = mg
               = (46 kg)(9.8 m/s2)
               = 450.8 N   (downward)
therefore , net force F = F' - w
                               F' = F + w
                                   = 1932 N + 450.8 N
                                   = 2382.8 N

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