Projectile motion including drag


by villiami
Tags: drag, including, motion, projectile
villiami
villiami is offline
#1
Nov1-05, 03:44 AM
P: 27
I have been working with projectile motion, and I am just starting to add air friction (drag) into the equations. I've run into a bit of a wall in terms of the calculations, so any help would be appriciated.

For a projectile, F(drag)=-c.V^2, where c is a constant (which can be written in terms of cross-sectional area, etc.)

Therefore: Acc(drag)=-c.V^2/(mass)


When I write an expression for vertical velocity [V(t)] at a given time [t], I get:

V(t) = V(initial) - 9.8t - (c/m). INTEGRAL{ [V(t)]^2 }dt


I then look at this equation and have trouble writting V(t) without an integral (or derivative for that matter).
I showed the problem to a friend, who gave me a strange tangent function for V(t), which I can't quite get my head around, as there is no mention of angles at this stage.


Maybe I'm on the wrong track, or maybe my calculus skills aren't quite up to scratch. Eventually I want to create a model for the projectile's motion (in terms of x and y), but first I need to get this part right.
Thanks for any help.
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dextercioby
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#2
Nov1-05, 04:01 AM
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Do you know how to separate variables in a differential equation?

Daniel.
villiami
villiami is offline
#3
Nov1-05, 04:07 AM
P: 27
Yes I realised that this is what I need to do to solve it (this resulted in an arctan funtion, which when rearranged gave me a tangent function for velocity). However, I was unsure how the tangent came out, and when I graphed the resulting function it did not look right. That's why I'm looking for some pointers.
Thanks,
villiami

dextercioby
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#4
Nov1-05, 04:13 AM
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Projectile motion including drag


This is your ODE

[tex] \frac{dv}{dt}=g -c \cdot v^{2} [/tex]

and you need to find [itex] v(t) [/itex] maybe with a initial condition giving the speed at t=0.

Daniel.
villiami
villiami is offline
#5
Nov1-05, 04:31 AM
P: 27
This then gives me [V is used instead of V(t)]:
dV/(g-cV^2)=dt
dV/[c(g/c-V^2)]=dt
INTEGRAL{1/c(g/c-V^2)}dV=INTEGRAL.dt
(1/sqrt[-g/c]).arctan(-Vc/g) = -cx + k; k is another constant
which eventually turns into:
V = (-g/c).tan{t.sqrt[-gc] + k.sqrt[-g/c]}
this doesn't seem to make sense when I draw the graph (I expected a curve with an asymptote at some terminal velocity)
Thanks,
Villiami
PS: how can I insert a formula? it takes me ages to write them in manually
dextercioby
dextercioby is offline
#6
Nov1-05, 04:45 AM
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P: 11,866
We use Latex, basically write the code and let the compiler do the rest.

Daniel.


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