Question

A parachutist whose mass is 75kg drops from a helicopter hovering 4000m above the ground and falls toward the earth under the influence of gravity. Assume the gravitational force is constant. Assume also that the force due to the air resistance is proportional to the velocity of the parachutist with the proportionality constant b_1 = 15N sec/m when the chute is close and b_2 = 105N sec/m when the chute is open. If the chute does open 1 min after the parachutist leaves the helicopter after how many seconds will he hit the ground?

Answer 1

Answer:

the parachutist hits the ground at t= 211.43 sec

Step-by-step explanation:

from Newton's second law:

Force = m*g - b*v = m*dv/dt

∫dt = ∫dv/(g - b/m*v) = - (m/b) ln (g - b/m*v) +C

t = - (m/b) ln (g - b/m*v) +C

assuming that he starts at rest (when he jumps) for t =0 → v=0

0 = - (m/b) ln (g - b/m*0) +C → C=(m/b) ln g

thus

t = (m/b) ln [g/(g - b/m*v)]

e^(b/m*t) = g/(g - b/m*v)

e^-(b/m*t) = 1- b/(m*g)*v

v= m*g/b [1-e^-(b/m*t) ]

then

dh/dt=v

∫dh = ∫v * dt = v=  ∫ m*g/b [1-e^-(b/m*t) ] dt =  m*g/b*[t + m/b *e^-(b/m*t)]

h = m*g/b*[t + m/b *e^-(b/m*t)]

then

a) before it opens b₁= 15N sec/m , thus

h = 75*9.8/15*[ 60 + 75/15 *e^-(15/75*60)] = 2940 m

b) after if opens  b₂= 105N sec/m , thus for the remaining meters

4000 - 2940 = 75*9.8/105*[ t + 75/105 *e^-(105/75*t)] =

since the equation is not linear we can neglect the exponencial term ( assume that reaches fast the terminal velocity since b is high), then

4000 - 2940 = 75*9.8/105* t

t= 151.43 sec

then to prove our assumption

additional distance neglected = 75/105 *e^-(105/75*151.43 )] = 2,36*10⁻³⁷ m ≈ 0

then our assumption is right

thus total time is

t total =  60 sec + 151.43 sec = 211.43 sec

therefore the parachutist hits the ground at t= 211.43 sec