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.

Question

Mathematics

Joseph Sheppard

13 July, 07:23

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.

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Yareli Morrison · Answer 1

For the presented problem, the solution would be v (0) = 0 v (0) = 0 is v (t) - mgb = e - b / m ⋅ t (v 0 - mgb) ⟺ v (t) = mgb (1 - e - b / m ⋅ t) ≈ g ⋅ t - gb 2 m ⋅ t 2, with the following given,

m = 180[lb]=81.6[kg]

g = 9.81[m/s 2 ]

b = 0.75[kg ⋅ m/s 2 ⋅ 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 [ f t / s ]. I am hoping that this answer has satisfied your query about this specific question.