restart: with(DEtools):Section 2.3 Problem #23A sky diver weighing LyomSSJtRzYiIiIiSSJnR0YlRiYiJCE9lbs. (including equipment) falls vertically downward from an altitudeof LyZJInhHNiI2IyIiISIlK10= ft. and opens the parachute after 10 seconds of free fall. Assume that the force of air
resistance is LyomSSNtdUc2IiIiIi1JJGFic0clKnByb3RlY3RlZEc2Iy1JInZHRiU2I0kidEdGJUYmLCRGJyQiI3YhIiM= when the parachute is closed and LyomSSNtdUc2IiIiIi1JJGFic0clKnByb3RlY3RlZEc2Iy1JInZHRiU2I0kidEdGJUYmLCRGJyIjNw== when theparachute is open, where LUkidkc2IjYjSSJ0R0Yk 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 Jkkidkc2IjYjSSJMR0Yk 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 LUkieEc2IjYjSSJ0R0Yk the altitude of the sky diver in feet, SSJ0RzYi
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
Ly1JJGFic0clKnByb3RlY3RlZEc2Iy1JInZHNiI2I0kidEdGKSwkRichIiI= also notice that air resistance "pulls up" as the diver falls so (as long as LUkiPkclKnByb3RlY3RlZEc2JEkjbXVHNiIiIiE=) we havethe force due to air resistance is LCQqJkkjbXVHNiIiIiItSSJ2R0YlNiNJInRHRiVGJiEiIg==.
Remembering that L0kiRkc2IkkjbWFHRiQ= and that Ly1JImFHNiI2I0kidEdGJS1JJWRpZmZHJSpwcm90ZWN0ZWRHNiQtSSJ2R0YlRiZGJw==, we get the following differential equation (SSJnRzYi is the acceleration due to gravity):
g := 32; m := 180/g;theEquation := m*diff(v(t),t) = -mu * v(t) - m*g;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 L0kjbXVHNiIkIiN2ISIj for the fall without parachute.mu := 0.75;closedVelocity := simplify(dsolve({theEquation,v(0)=0}));(a) What is the sky diver's velocity when the chute opens (at L0kidEc2IiIjNQ== seconds)?v10 := subs(t=10,rhs(closedVelocity));evalf(abs(v10));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 Ly1JInhHNiI2IyIiISIlK10=.closedPositionEq := diff(x(t),t) = rhs(closedVelocity);closedPosition := simplify(dsolve({closedPositionEq,x(0)=5000}));(b) How far has the sky diver fallen when the chute opens?x10 := subs(t=10,rhs(closedPosition));evalf(5000-x10);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 Ly1JInZHNiI2IyIjNUkkdjEwR0Yl and Ly1JInhHNiI2IyIjNUkkeDEwR0Yl.Remember that L0kjbXVHNiIiIzc= for this part of the problem.mu := 12;theEquation;openVelocity := simplify(dsolve({theEquation,v(10)=v10}));openPositionEq := diff(x(t),t) = rhs(openVelocity);openPosition := simplify(dsolve({openPositionEq,x(10)=x10}));(c) What is the limiting velocity Jkkidkc2IjYjSSJMR0Yk for the sky diver after the chute has opened?v_L := limit(rhs(openVelocity),t=infinity);
(d) Determine how long the sky diver is in the air after his parachute opens.
hitsGround := simplify(solve(rhs(openPosition)=0,t));evalf(hitsGround) - 10;(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);