Problem Set D
Problem 7
Ÿ (a)
The equation of motion during free fall is my''= -32m+(.005)(y')^2, or y''+32 - (4/4875) (y')^2=0, where m=the
mass of the jumper. The upward direction is positive and y=0 at ground level, so the initial conditions are
y(0)=1000 and y'(0)=0. We now solve this problem with NDSolve.
Clear[sol1,ya1]
sol1 = NDSolve[{y''[t]+32-(4/4875)y'[t]^2==0,y[0]==1000,
y'[0]==0},y[t],{t,0,5}]
88y@tD ® InterpolatingFunction@880., 5.<<, <>D@tD<<
ya1[t_]=y[t]/.First[sol1]
InterpolatingFunction@880., 5.<<, <>D@tD
ya1[5]
637.361
ya1'[5]
- 132.254
So the height when the chute opens is 637.361 ft. and the velocity is -132.254 ft./sec. To analyze the second phase
of the fall we use NDSolve as follows.
Clear[sol2,ya2]
sol2 = NDSolve[{y''[t]+32-(192/1950)y'[t]^2==0,
y[5]==637.361,y'[5]==-132.254},y[t],{t,5,45}]
88y@tD ® InterpolatingFunction@885., 45.<<, <>D@tD<<
ya2[t_]=y[t]/.First[sol2]
InterpolatingFunction@885., 45.<<, <>D@tD
Note that we have used a fairly large time interval in order to get beyond the time of impact.
Plot[ya2[t],{t,5,45}]
600
500
400
300
200
100
20
-100
30
40
2
PSD07.nb
ya2[39]
9.91972
ya2[40]
- 8.10803
We see that the time of impact is between 39 and 40 sec. To find an accurate value we use FindRoot.
FindRoot[ya2[t],{t,40}]
8t ® 39.5502<
To find the velocity of impact we evaluate ya2'[39.5502].
ya2'[39.5502]
- 18.0278
So the velocity of impact is -18.0278 ft./sec. or -12.2917 miles/hour.
Ÿ (b)
The first order equation in v is v'=-32+(192/1950)v^2. Thus the critical point is
Sqrt@32 * 1950  192D  N
18.0278
This is considerably less than the velocity at the time the chute opens, but it is equal to the velocity of impact. More
precisely, the velocity at the moment of impact might well be somewhat bigger than the terminal velocity, but to six
digits it is the same. So the jumper (essentially) reached terminal velocity before landing.
Ÿ (c)
A safe velocity is one which is greater tha 1.05 times the terminal velocity, thus
1.05*(-18.0278)
- 18.9292
Therefore as long as the diver's velocity is not (algebraically) less than -18.9292
when the ground is reached, safety is assured. It is enough to run the preceding model
and see at what time the velocity was -18.9292; then compute the distance fallen until
that point.
FindRoot[ya2'[t]==-18.9292,{t,5.5}]
8t ® 5.96876<
In fact, in only 5.96877 seconds after the parachute opens, the diver effectively attains
the safe velocity. The height at that time is
ya2[5.96877]
605.649
Therefore, the diver is safe as long as the jump is from a height at least
1000-605.649=394.351ft. from the ground.