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A skydiver weighing 180lbs (including equipment) falls vertically downward from an altitude of 5000 ft and opens the parachute after 10 s of free fall. Assume that the force of air resistance, which is directed opposite to the velocity, is of magnitude.75|v| when the parachute is closed and is of magnitude 12|v| when the parachute is open, where velocity v is measured in ft/s. Find the speed of the skydiver when the parachute opens.

User Hfmanson
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For the presented problem, the solution would be v(0)=0v(0)=0 isv(t)−mgb=eb/mt(v0mgb)v(t)=mgb(1−eb/mt)gtgb2mt2, with the following given,
m
=180[lb]=81.6[kg]

g=9.81[m/s2]

b=0.75[kgm/s2s/ft]=0.2286[kg/s]

The solution that the friction provides is v(t)=3501.7[m/s](1−e−0.00280[1/s]t), where I get 96.69[m/s]=317.2[ft/s]. I am hoping that this answer has satisfied your query about this specific question.

User Narayanpatra
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