# 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!

Thank you Relevance

∑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

Integrate:

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

Graph:

https://www.desmos.com/calculator/kq9r7hjmcl

The skydiver's velocity is 34.8 ft/s downward.

• Just a matter of convention.  Your method works if you define down as positive. But I defined it as negative.

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• 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.

=>

y"=g-ky'/m.

g=gravitational acceleration.

when the parachute is closed, k=0.8

When the parachute is open, k=5.4

are given.

Solving the diff. equ., get

dy'/[y'-w/(gk)]=-kgdt/w

=>

ln[y'-w/(gk)]=-kgt/w+lnC

=>

y'=Ce^(-kgt/w)+w/(gk)

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

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