How Do You Calculate Projectile Motion with Air Drag?

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The discussion revolves around calculating projectile motion while incorporating air drag into the equations. The main equation presented is F(drag) = -c.V^2, leading to an acceleration expression that includes drag. Participants discuss the difficulty of expressing vertical velocity V(t) without using integrals or derivatives, and the emergence of a tangent function in the solution process. A differential equation is formulated, and through integration, a relationship involving arctan is derived, although concerns are raised about the resulting graph not matching expected behavior, particularly regarding terminal velocity. The conversation highlights the challenges of solving ordinary differential equations in the context of projectile motion with air resistance.
villiami
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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.
 
Last edited:
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Do you know how to separate variables in a differential equation?

Daniel.
 
Yes I realized 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
 
This is your ODE

\frac{dv}{dt}=g -c \cdot v^{2}

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

Daniel.
 
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
 
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We use Latex, basically write the code and let the compiler do the rest.

Daniel.
 
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