>	restart: with(DEtools):

 

Section 2.3 Problem #23 

A sky diver weighing  lbs. (including equipment) falls vertically downward from an altitude 

of ft. and opens the parachute after 10 seconds of free fall. Assume that the force of air
resistance is when the parachute is closed and when the 

parachute is open, where is the velocity measured in feet per second. 

(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 after the parachute opens? (Terminal velocity) 

(d) Determine how long the sky diver is in the air after the parachute opens. 

(e) Plot the graph of velocity versus time from the beginning of the fall until the sky diver reaches the ground.


We will make "up" positive and "down" negative. We will call the altitude of the sky diver in feet,
seconds after the start of his descent. 

Notice that since the sky diver is falling during the entire problem, his velocity is always negative so that
also notice that air resistance "pulls up" as the diver falls so (as long as ) we have 

the force due to air resistance is .
 

Remembering that and that , we get the following differential equation ( is the acceleration due to gravity):
 

>	g := 32; m := 180/g;

 

 

	(1)

 

>	theEquation := m*diff(v(t),t) = -mu * v(t) - m*g;

 

(2)

 

We solve the differential equation governing the descent with the parachute closed. We will assume that
the sky diver's initial velocity is 0. Recall that for the fall without parachute. 

>	mu := 0.75;

 

(3)

 

>	closedVelocity := simplify(dsolve({theEquation,v(0)=0}));

 

(4)

 

(a) What is the sky diver's velocity when the chute opens (at seconds)? 

>	v10 := subs(t=10,rhs(closedVelocity));

 

(5)

 

>	evalf(abs(v10));

 

(6)

 

Now that we know the velocity (for the chute closed), we can integrate and find the position function.
However, we must remember to plug in the initial position . 

>	closedPositionEq := diff(x(t),t) = rhs(closedVelocity);

 

(7)

 

>	closedPosition := simplify(dsolve({closedPositionEq,x(0)=5000}));

 

(8)

 

(b) How far has the sky diver fallen when the chute opens? 

>	x10 := subs(t=10,rhs(closedPosition));

 

(9)

 

>	evalf(5000-x10);

 

(10)

 

Next, we must find the velocity and position of the sky diver after his chute has opened. 

We have all ready calculated the initial conditions and . 

Remember that for this part of the problem. 

>

mu := 12;

 

(11)

 

>	theEquation;

 

(12)

 

>	openVelocity := simplify(dsolve({theEquation,v(10)=v10}));

 

(13)

 

>	openPositionEq := diff(x(t),t) = rhs(openVelocity);

 

(14)

 

>	openPosition := simplify(dsolve({openPositionEq,x(10)=x10}));

 

(15)

 

(c) What is the limiting velocity for the sky diver after the chute has opened? 

>	v_L := limit(rhs(openVelocity),t=infinity);

 

(16)

 

(d) Determine how long the sky diver is in the air after his parachute opens.
 

>	hitsGround := simplify(solve(rhs(openPosition)=0,t));

 

(17)

 

>	evalf(hitsGround) - 10;

 

(18)

 

(e) Graph the sky divers velocity function (we will also plot his position function). 

>	plotVClosed := plot(rhs(closedVelocity),t=0..10,numpoints=5000): plotVOpen := plot(rhs(openVelocity),t=10..hitsGround,numpoints=5000): plots[display](plotVClosed,plotVOpen);

 

>	plotXClosed := plot(rhs(closedPosition),t=0..10,numpoints=5000): plotXOpen := plot(rhs(openPosition),t=10..hitsGround,numpoints=5000): plots[display](plotXClosed,plotXOpen);