freefall: A parachutist of mass 60 kg free-falls from an....

mathstresser

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A parachutist of mass 60 kg free-falls from an airplane at an altitude of 5000 meters. He is subjected to an air resistance force that is proportional to his speed. Assume the constatnt of proportionality is 10 (kg/sec). Find and solve the differential equation governing the altitude of the parachuter at time t seconds after the start of his free-fall. Assuming he does not deploy his parachute, find his limiting velocity and how much time will elapse before he hits the ground.

Air resistance:

r= kv, k =10

terminal velocity= -mg/r

-(9.8)(60)/(10)

-58.8

How do I find the altitude and the equation for altitude? How do I find the time?
 
forces acting on the parachutist ...

mg downward and kv upward

as a scalar equation (strictly working as though velocity, acceleration, and displacement are all positive) ...

F<sub>net</sub> = mg - kv

ma = mg - kv

a = g - (k/m)v

dv/dt = g(1 - [k/(mg)]v)

let b = k/(mg) ...

dv/dt = g(1 - bv)

dv/(1 - bv) = g dt

-b/(1 - bv) dv = -bg dt

ln|1 - bv| = -bgt + C<sub>1</sub>

1 - bv = C<sub>2</sub>e<sup>-bgt</sup>

v = (1/b)(1 - C<sub>2</sub>e<sup>-bgt</sup>)

at t = 0, v = 0 ... C<sub>2</sub> = 1

v = (1/b)(1 - e<sup>-bgt</sup>)

v = (mg/k)(1 - e<sup>-kt/m</sup>)

lim{t -> inf} v(t) = mg/k

This result can also be confirmed from the initial differential equation a = g - (k/m)v. When the parachutist reaches terminal speed, a = 0 ... g = (k/m)v ... v = mg/k.

v = (mg/k)(1 - e<sup>-kt/m</sup>)

dx/dt = (mg/k)(1 - e<sup>-kt/m</sup>)

x = (mg/k)[t + (m/k)e<sup>-kt/m</sup>] + C

at t = 0, x = 0 (distance from original position) ... C = -352.8

5000 = 58.8(t + 6e<sup>-t/6</sup>) - 352.8

at x = 5000 m (the ground), t = approx 91 sec
 
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