Question

According to a newspaper account, a paratrooper survived a training jump from 1214 ft when his parachute failed to open but provided some resistance by flapping in the wind. Allegedly he hit the ground at 97.5 mih after falling for 7 seconds. To test the accuracy of this account, you should first find the drag coefficient p, assuming a terminal velocity of 97.5 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-l Next, find the distance D fallen in 7 seconds is the newspaper account reasonable? Enter Y or N

Answer 1

We take downward direction as positive in the problem

now,

acceleration is given by   

dv dt

according to the problem

or

dt

The integrating factor is therefore

opt

(eptu) = get dt

ot et C

or

eq-1

we know that the terminal velocity is 97.5 m/h or 143 ft/s

as time tends to infinity    

by plugging all the values we get

or

(as g=32 ft/s² is given.)

thus we get

Part-b

Again using equation -1 in y direction displacement over time we get

or

eq-2   (integrating w.r.t time)

we know that at   $t=0$

plugging these values in eq-2 we get

or

which gives us

thus the eq-2 will become

For the distance travelled in 7 seconds we have to put t=7 in above equation, therefore

we get

Hence the distance fallen in 7 seconds is