Question

Suppose a small cannonball weighing 16 pounds is shot vertically upward, with an initial velocity
v0 = 300 ft/s.
The answer to the question "How high does the cannonball go?" depends on whether we take air resistance into account. If air resistance is ignored and the positive direction is upward, then a model for the state of the cannonball is given by
d2s/dt2 = −g
(equation (12) of Section 1.3). Since
ds/dt = v(t)
the last differential equation is the same as
dv/dt = −g,
where we take
g = 32 ft/s2.
If air resistance is incorporated into the model, it stands to reason that the maximum height attained by the cannonball must be less than if air resistance is ignored.
(a) Assume air resistance is proportional to instantaneous velocity. If the positive direction is upward, a model for the state of the cannonball is given by
m dv/dt = −mg − kv,
where m is the mass of the cannonball and
k > 0
is a constant of proportionality. Suppose
k = 0.0025
and find the velocity
v(t)
of the cannonball at time t.

Answer 1

m dv/dt = -mg -kv
v' + (k/m)v = -g

use the integrating factor e^(kt/m) and you have

(e^(kt/m) v)' = -ge^(kt/m)
Now integrate and you have

e^(kt/m) v = -g(m/k)e^(kt/m) + c
v = -gm/k + c*e^(-kt/m)

Now just plug in your values

Answer 2

ok will try to use feet instead of meters
mass = weight/g
m = 16 pounds/32 = .5

m dv/dt = - m g - k v

m dv/dt = -32 m - k v

dv/dt = -32 -(.0025/.5) v

dv/dt = -32 - .005 v

try v = - b + a e^-c t)
where a c and b are constants
then
dv/dt = -b - a c e^-ct
so
-32-.005(-b+ae^-ct) = -b -ac e^-ct
-32 +.005b -.005ae^-ct = -b -ace^-ct
32 = 1.005 b
b = 31.8
.005 = c
so
v = -31.8 + a e^-.005 t
but at t = 0, v = 300
300 = -31.8 + a e^0
a = 331.8

so v =-31.8 + 331.8 e^-.005 t

Answer 3

I worked it out without the integrating factor. I end up with
v=-m/k Ce^-(kt/m) - mg/k

In webassign, if I do the practice problem and get the wrong answer I can ask for the correct one. It gives the solution as 6700e^-0.005t - 6400
I'm not coming up with that.

I think my solution via separable DE is the same as yours via integrating factor....

Answer 4

That's the problem. I didn't change weight to mass.

Thanks for the help!