4) A sky diver weighing 120 kg (including equipment) falls vertically downward f

Question

4) A sky diver weighing 120 kg (including equipment) falls vertically downward from an altitude of 5000 m, and opens the parachute after 40 sec of free fall. Assume that the force of air resistance is ?10v when the parachute is closed and ?120v when the parachute is open, where the velocity v is measured in m/sec.
(a) Find the speed of the sky diver when the parachute opens.
(b) Find the distance fallen before the parachute opens.
(c) What is the limiting velocity vL after the parachute opens?
(d) Determine how long the sky diver is in the air after the parachute opens.

Explanation / Answer

m*dv/dt = mg - v

Before the paracute open :

= 10

dv/dt + v/12 = 9.8

Integrating factor = et/12

Solution is:

et/12v(t) = 9.8 integrate( et/12)

et/12v(t) = 9.8*12* et/12+C

V(0)=0

C=9.8*12

V(t) = 9.8*12-9.8*12* e-t/12

=117.6-117.6* e-t/12

(a)

V(40) = 117.6-117.6* e-40/12

= 113.405 m/s

(b)

We denote by s(t) the displacement at time t in the downward (positive) direction, that is,the height is 5000 s(t). To find s(t) we integrate the velocity

s(t) = Integrate(V(t))

=117.6*t+1411.2*e-t/12+C

S(0) = 0 implies C= - 1411.2

s(t) =117.6*t+1411.2*e-t/12-1411.2

S(40) = 117.6*40+1411.2*e-40/12-1411.2

= 3343.143 m

(c)

After the parachute opens

Now, = 120

dv/dt + v= 9.8

Integrating factor = et

Solution is:

etv(t) = 9.8 integrate( et)

etv(t) = 9.8et+C

V(0) = 113.405 m/s

C=103.605

V(t) =  9.8+103.605*e-t

The limiting velocity (as t) is

V(L) = 9.8 m/s

(d)

s(t) = integrate(V(t))

=9.8*t - 103.605*e-t+C

S(0) = 3343.143 m

S(t)=9.8*t - 103.605*e-t+3446.748

Finally,

S(t) =5000

5000=9.8*t - 103.605*e-t+3446.748

-9.8*t +103.605*e-t+1553.252 =0

solving,we gat

t=158.495 sec

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