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 1152 ft when his parachute failed to open but provided some resistance by flapping in the wind. Allegedly he hit the ground at 104.25 mi/h after falling for 8.5 seconds. To test the accuracy of this account, you should first find the drag coefficient, assuming a terminal velocity of 104.25 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/sec^2.

drag coefficient= _____sec^-1

Next, find the distance D fallen in 8.5 seconds.

D= _____ft

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Explanation / Answer

The CF is Aexp(-ct) + B and the PI is gt/c so solution is s = Aexp(-ct) + gt/c + B At t=0, s=0, ds/dt=0 so 0 = A+B and 0 = -cA+g/c ? A = -B = g/c² ? s(t) = (g/c²){ exp(-ct) - 1 } + gt/c Assuming he reached his terminal velocity when he hit the ground c.ds/dt = g 104.25 mph = 104.25 x 1609.34 / 3600 = 46.60m/s. This gives c = 9.81/46.60 = 0.2103 s?¹ . Without mass you cannot work out k. Using this value s(t) = 210.85{ exp(-0.2103t) - 1 } + 46.60t So s(6) = 210.85{ exp(-6.5x0.2103) - 1 } + 46.60x6.5 = 187.65 m This is very different to 1152 ft. I believe my arithmetic to be correct. If you put c=0 you get fall with no resistance and s=½ gt². Even this only gives 146.58 m.

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