Calculus 2 homework?
A 180-lb skydiver jumps out of an airplane (with zero initial velocity). Assume that k=0.8 lb-s/ft with a closed parachute and k=5.4 lb-s/ft with an open parachute. What is the skydiver's velocity at t=25 s if the parachute opens after 20 s of free fall?
I do not have any work to really provide. I'm confused on how to set this question up. Any help would be great!
- Some BodyLv 72 years agoFavorite Answer
Start with Newton's second law:
∑F = ma
The forces acting on the skydiver are gravity and drag:
ma = -kv − mg
m dv/dt = -kv − mg
-m/k dv/dt = v + mg/k
Separate the variables:
dv / (v + mg/k) = -k/m dt
ln |v + mg/k| = -k/m t + C
v + mg/k = e^(-k/m t + C)
v + mg/k = Ce^(-k/m t)
v = -mg/k + Ce^(-k/m t)
At time t=0, v=0:
0 = -mg/k + C
C = mg/k
v = mg/k (-1 + e^(-k/m t))
mg = 180 lb, g = 32.2 ft/s², k = 0.8 lb-s/ft and t = 20s:
v = (180 lb / 0.8 lb-s/ft) (-1 + e^(-0.8 lb-s/ft / (180 lb / 32.2 ft/s²) * 20s))
v = -212 ft/s
Now the skydiver opens his parachute. At time t=20, v = -212 ft/s:
-212 = -mg/k + Ce^(-k/m * 20)
C = (-212 + mg/k) e^(20k/m)
v = -mg/k + (-212 + mg/k) e^(k/m (20 − t))
mg = 180 lb, g = 32.2 ft/s, k = 5.4 lb-s/ft, and t = 25s:
v = (-180 lb / 5.4 lb-s/ft) + (-212 ft/s + 180 lb / 5.4 lb-s/ft) e^(5.4 lb-s/ft / (180 lb / 32.2 ft/s²) * (20s − 25s))
v = -34.8 ft/s
The skydiver's velocity is 34.8 ft/s downward.
- PinkgreenLv 72 years ago
The resistant by air is -ky' by experiment for a parachute=> the equation of motion
at time t is
my"=w-ky', where w=the weight of diver, y"=the downward acc., y'=downward speed,
k=constant, m=the mass of the diver.
when the parachute is closed, k=0.8
When the parachute is open, k=5.4
Solving the diff. equ., get
During the 1st 20 s, the parachute is closed,
taking k=0.8 & work out C=C1 by means of the
initial condition that y'=0 as t=0.
When C1 is known, find y'(20)=u which is the
initial downward speed foe the later 5 s.
After 20 s, the parachute is open, now take k=5.4.
Find C=C2 again by inserting t=0 & Y'=u. The
diff.equation now becomes
y'=C2e^(-5.4gt/w)+w/(5.4g), then work out
y'(5)=C2e^(-5.4g(5)/w)+w/(5.4g) as the answer.
In this problem, whether 180 lb is w or m had not been