A skydiver is at an altitude of 1,529 m. After 11.0 seconds of free fall, he ope

Question

A skydiver is at an altitude of 1,529 m. After 11.0 seconds of free fall, he opens his parachute and finds that the air resistance, FD, is given by the formula FD = bv, where b is a constant and v is the velocity. If b = 100, and the mass of the skydiver is 88.0 kg, first set up differential equations for the velocity and the position, and then find the following. (Assume air resistance is negligible prior to the skydiver opening his parachute.)

(a) the speed (m/s) of the skydiver when the parachute opens m/s

(b) the distance (m) fallen before the parachute opens m

(c) the terminal velocity (m/s) after the parachute opens (find the limiting velocity) (Enter the magnitude.) m/s

(d) the time (in s) the skydiver is in the air after the parachute opens

Explanation / Answer

a )

using v= u + a t

v = 0 + 9.8 X 11

v = 107.8 m/s

b )

s = ut + 1/2 a t2

s = 0 + 0.5 X 9.8 X 112

s = 592.9 m

c )

FD = b v

m g = b v

v = m g / b

v = 88 X 9.8 / 100

v = 8.624 m/s

d )

time = distance / velocity

time = 1529 - 592.9 / 8.624

time = 108.5459sec

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