Kinematics of pumpkin shot out of air cannon

[tb.MSU]

A pumpkin of mass 5kg is shot out of an air cannon at an elevation angle of 45 degrees with an initial velocity of 54m/s. It lands 142 m away. Assuming the force due to air resistance was F=-kmv, find the value of k.

I need some help with this problem. I'm not quite sure how to put in the angle. All i really know is:

F = ma = m(dv/dt) = -kmv ... after integration

ln v = -kt + C1 v(t=0)=Vo C1 = ln Vo
thus:
v = Vo(e^-kt)

i know this is completely off though because no where is there a sin or cos for me to input the angle..


any help would be much appreciated thanks!

ps. I'm a second year student at Michigan State. Astrophysics major.. gone through Calc 4.

tb

 

[dextercioby]

U forgot about the gravity force and the fact that the 2-nd law of Newton is a vector equation.

Daniel.

 

[tb.MSU]

i'm actually starting to get somewhere now...

x(t=0) = 0 = y(t=0)
.
x(t=0) = Vo cos theta
.
y(t=0) = Vo sin theta
.. .
mx = -kmx
.. .
my = -kmy - mg


and I'm basically solving from there... i think i'll be okay on this question now, but any pushes while I'm on the swing will be w00teriffic. :)

 

[dextercioby]

Okay,post any "accidents" along the way.

Daniel.

 

[tb.MSU]

okay figured it out

i ended up having to set separate variables for time and also had to find the projectile range with and without air resistance.

this put me in an expansion parameter or coupling constant. For short, i used the "perturbation method" to come to a final answer of .257s^-1

R` = R (1 - ((4kVo sin theta)/3g))
where R = (Vo^2/g)sin 2 theta



Anyway,
I just found this page surprisingly. I was reading through a lot of other posts and noticed that there are several intelligent people posting and commenting on this site, so i figured i'd try and join up in hopes to better myself and others is possible.

Cheers,

tb

(GO STATE!)