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

Answer:

F = 3482.9 N

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

$a = \frac{v_f^2 - v_i^2}{2d}$

distance moved by the parachutist is given as

$d = 120 ft = 36.576 m$

now we have

$a = \frac{6.675^2 - 53.4^2}{2(36.576)}$

$a = - 38.4m/s^2$

Now the mass of parachutist is given as

$m = 200 lb = 90.7 kg$

now we have

$F = ma$

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