Question

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.

Answer 1

For the presented problem, the solution would be v(0)=0v(0)=0 isv(t)−mgb=e−b/m⋅t(v0−mgb)⟺v(t)=mgb(1−e−b/m⋅t)≈g⋅t−gb2m⋅t2, with the following given,
m=180[lb]=81.6[kg] 

g=9.81[m/s2]

b=0.75[kg⋅m/s2⋅s/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.