Question #70331

A person falls from rest at a height of five meters. Their mass is 70kg. When they land the impact has a duration of 0.1 seconds. What forces was developed on impact
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Expert's answer

2017-10-02T12:32:07-0400

Answer on Question #70331-Physics-Other

A person falls from rest at a height of five meters. Their mass is 70kg. When they land the impact has a duration of 0.1 seconds. What forces was developed on impact

Solution

The velocity before impact is


v=2gh.v = \sqrt {2 g h}.


The force is


F=mvt=m2ght=702(9.8)50.1=6930N.F = \frac {m v}{t} = \frac {m \sqrt {2 g h}}{t} = \frac {7 0 \sqrt {2 (9 . 8) 5}}{0 . 1} = 6 9 3 0 N.


Answer: 6930 N.

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Comments

Assignment Expert
29.09.17, 17:34

Dear visitor, please use panel for submitting new questions

Shamsuddoha
29.09.17, 17:28

Person's weight = 70 kg. Velocity initial =0m/s and Velocity final = 5m/0.1s =50m/s. Use acceleration: a=change of V / change of time = 50-0/.1-0 = 500m/s^2. Now Force = mass*a = (70*500)kg*m/s^2 =3500N

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