A skydiver weighs 125 pounds, and their equipment weighs another 35 pounds. Afte

Question

A skydiver weighs 125 pounds, and their equipment weighs another 35 pounds. After jumping out of the plane at 15,000 feet, they wait 15 seconds then open their parachute.

Given this model for air resistance: m(dv/dt) = mg-kv where k > 0 is a constant of proportionality.

Assume that the constant of proportionality has the value k=.5 during freefall and k=10 after the parachute is opened. Assume that the skydiver's initial velocity upon leaving the plane is zero. What is their velocity and how far have they travelled 20 seconds after leaving the plane? How does their velocity at 20 seconds compare with her terminal velocity?

Explanation / Answer

m dv/dt = mg - kv ..................... mass * acceleration = net force
dv/dt + k/m v = g
d/dt [ e^(kt/m) v ] = g e^(kt/m)
e^(kt/m) v = gm/k e^(kt/m) + C
v = C e^(-kt/m) + gm/k
V = C + gm/k ....... C = V - gm/k
v = (V - gm/k) e^(-kt/m) + gm/k
v(20) = (0+32.174(160)/0.5))*e^(-0.5*20/160) - 32.174*160/0.5

Answer: v(20) 623.8 ft/s

x = v dt
= (V - gm/k) e^(-kt/m) dt + gm/k dt
= (g(m/k)² - V m/k) e^(-kt/m) + gm/k t + C
X = (g(m/k)² - V m/k) + C .... C = X - (g(m/k)² - V m/k)
x = (g(m/k)² - V m/k) (e^(-kt/m) - 1) + gm/k t + X
x(20) = (32.174(160/0.5)²)(e^(-0.5(20)/160) - 1) + 32.174(160)(20)/0.5

Answer: x(20) = 6302.8 ft

lim (t ) (V - gm/k) e^(-kt/m) + gm/k
= gm/k
= 32.174(160)/10
= 514.8 ft/s

Answer: terminal velocity = 514.8 ft/s

x = (g(m/k)² - V m/k) (e^(-kt/m) - 1) + gm/k t + X
15000 = (32.174(160/10)² - (623.8)(160)/10) (e^(-10t/160) - 1) + 32.174(160)/10 t + 6302.8
t 14.85
reach the ground at 20 + 14.85 = 34.85 s

Answer: 34.85 s

intc_escapee · 6 years ago

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