Question

A parachutist who weighs 200lbs is falling at 120 miles/hour when his parachute opens. His speed is reduced to 15 miles/hour in a vertical distance of 120ft. What force did the parachute exert on the jumper?

Answer 1

F = 3482.9 N

Step-by-step explanation:

Change in velocity of the Parachutist is given as

`v_f = 15 mph = 6.675 m/s`

`v_i = 120 mph = 53.4 m/s`

now it is given as

`\Delta v = v_f - v_i`

`\Delta v = 120 - 15 = 105 mph`

`\Delta v = 46.7 m/s`

now the acceleration of the parachutist is given as

`a  = (v_f^2 - v_i^2)/(2d)`

distance moved by the parachutist is given as

`d = 120 ft = 36.576 m`

now we have

`a = (6.675^2 - 53.4^2)/(2(36.576))`

`a = - 38.4m/s^2`

Now the mass of parachutist is given as

`m = 200 lb = 90.7 kg`

now we have

`F = ma`

`F = (90.7 kg)(38.4) = 3482.9 N`