A parachutist of mass 70 kg jumps from a plane at an altitude of 32 km above the

Question

A parachutist of mass 70 kg jumps from a plane at an altitude of 32 km above the surface of the Earth. Unfortunately the parachute fails to open. (In the following parts, neglect horizontal motion an assume that the initial velocity is zero.)

A. Calculate the time of fall (accurate to 1 s) until ground impact, given no air resistance and a constant value of j.
B. Calculate the time of fall (accurate to 1 s) until ground impact, given constant g and a force of air resistance given by F(v) == c2v|v| where c2 is 0.5 in SI units for a falling man and is constant.

Explanation / Answer

part a)

initial velocity = 0

we know that s = ut + 0.5gt^2 (second law of motion)

32000 = 0t + 0.5*9.8*t^2

t = sqrt(32000/0.5*9.8) = 80.81 sec

part b)

Net force = mg - 0.5v|v| = mg -/+ 0.5v^2

thus net acceleration = g - 0.5v|v|/m

v = u + gt = gt (since u = 0)

32000 = 0.5*(g +0.5g^2t^2/m)t^2

32000 = 0.5*g(1 + 0.5*g*t^2*0.007)t^2

6530.61 = t^2 + 0.0343t^4

t^2 = 422.011 s

t = 20.54 s

0.007t^4 - t^2 + 6530.61 = 0

t = 20.543 s

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