Question

A skydiver is at an altitude of 1,525 m. After 11.0 seconds of free fall, he...

A skydiver is at an altitude of 1,525 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 86.0 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

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

c: the terminal velocity (m/s) after the parachute opens

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

Given: Altitude 1525 m, 11sec spent before the opening of the parachute, m= 86kg, b=100

Before opening the parachute:

ma= mg (only gravitational force acting downwards)

After opening the parachute:

ma= mg-bv (now drag force is acting upwards)

a= g- (b/m)v

dv/dt = g-(b/m)v

(a) The speed of the driver when the parachute opens:

v= u + at

v= 0+ 9.8x 11= 107.8 m/s

(b) s= ut + (1/2) at2

s= 0+ (1/2) x 9.8 x (11)2= 592.9 m

(c) Terminal velocity is attained when gravitational force becomes equal to drag force

mg=bv

v= mg/b= 86x9.8/100 = 8.43 m/s

(d) Time = distance/ velocity = (1525-592.9)/8.43 = 74.98 sec