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?

Respuesta :

Answer:

F = 3482.9 N

Explanation:

Change in velocity of the Parachutist is given as

[tex]v_f = 15 mph = 6.675 m/s[/tex]

[tex]v_i = 120 mph = 53.4 m/s[/tex]

now it is given as

[tex]\Delta v = v_f - v_i [/tex]

[tex]\Delta v = 120 - 15 = 105 mph[/tex]

[tex]\Delta v = 46.7 m/s[/tex]

now the acceleration of the parachutist is given as

[tex]a  = \frac{v_f^2 - v_i^2}{2d}[/tex]

distance moved by the parachutist is given as

[tex]d = 120 ft = 36.576 m[/tex]

now we have

[tex]a = \frac{6.675^2 - 53.4^2}{2(36.576)}[/tex]

[tex]a = - 38.4m/s^2[/tex]

Now the mass of parachutist is given as

[tex]m = 200 lb = 90.7 kg[/tex]

now we have

[tex]F = ma[/tex]

[tex]F = (90.7 kg)(38.4) = 3482.9 N[/tex]

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