According to a newspaper account, a paratrooper survived a training jump from 11

Question

According to a newspaper account, a paratrooper survived a training jump from 1190 ft when his parachute failed to open but provided some resistance by flapping in the wind. Allegedly he hit the ground at 96 mi/h after falling for 10 seconds. To test the accuracy of this account, you should first find the drag coefficient , assuming a terminal velocity of 96 mi/h and also that the resistance of the paratrooper falling through the air is proportional to his velocity. Remember that the accleration due to gravity near the earth's surface is 32 ft/sec2 sec Next, find the distance D fallen in 10 seconds. D= Is the newspaper account reasonable? Enter Y or N.

Explanation / Answer

Force acting on paratrooper are gravity (downwards) and air drag (upwards) . Therefore net acceleration is

a = g - pv , where p is the drag coefficient .

Given : terminal velocity vT = 96 mile/hr = 96 * 5280 ft/ 3600 sec = 140.8 ft/sec

a becomes zero when velocity reaching terminal velocity(vT)

Now a = 0 => g - pvT = 0 => p = g / vT = (32 ft/sec2) / (140.8 ft/sec) => p = 0.227 sec-1

a = dv/dt = g - p v

=> dv/g-pv = dt

Integrate both sides we get

=> -1/p * ( ln(g-pv))  = t + c (c is contant of integration)

=> g - pv = k*e-pt (k = e-pc)

At t = 0 , v = 0 => g - 0 = k*1 => k = g

therefore

g - pv = ge-pt

=> v = g(1-e-pt) / p

we know that v = dx/dt

=> dx/dt = g(1-e-pt) / p

=> dx = (g/p)*(1-e-pt) dt

integrate both sides

=> x = g/p*(t + e-pt / p) + c2 (where c2 is the constant of integration)

when t = 0 , c = 0 therefore 0 = g/p*(0+1/p)+c2 => c2 = -g/p2

Hence x = g/p*(t + e-pt / p) - g/p2 = g/p*(t + (e-pt -1 )/ p)

Now then t = 10 seconds , x = (9.8 / 0.227)*(10 +e-0.227*10 -1 / 0.227) = 140.8 *( 10 - 3.95) = 140.8 * 6.05 = 851.81 feet

D = 851.81 feet

Since distance covered is less than what claimed in 10seconds , therefore there must be some force due to parachute and hence the newspaper account is not reasonable [N]

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