Question

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, *F*_{D}, is given by the formula
*F*_{D} = −*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

Answer #1

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) at^{2}

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

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