A skydiver weighs 125 lbs and her parachute and other equipment combined weigh a

Question

A skydiver weighs 125 lbs and her parachute and other equipment combined weigh another 35 lbs. After exiting from a plane at an altitude of 15,000 feet, she waits 15 seconds and opens her parachute. Assume that the constant of proportionality in the model is m(dv/dt)=mg-kv where k=0.5 during the free fall and k=10 after release of the parachute. Assume that her initial velocity on leaving the plane is zero. What is her velocity and how far has she traveled after 20 seconds from leaving the plane? How does her velocity at 20 seconds compare with her terminal velocity? How long does it take her to reach the ground? [Hint: Think in terms of two distinct IVPs.]

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

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