Time of flight with air resistance

[Caspian]

I need to compute the time of flight of a projectile which is subject to air resistance.

Here's where I am so far in solving the problem:

F_{drag} = -B v^2
a = \frac{\partial v}{\partial t} = \frac{-B v^2}{m}

integrate and solve for v...

v = \sqrt[3]{\dfrac{1}{3 \dfrac{B}{m} t + C}}

I then plug this into the kinematics equation for position:

y = \frac{1}{2} a t^2 + v_0 t (y_0 is taken to be 0)
y = \frac{1}{2} (g - \frac{F_{drag}}{m}) t^2 + v_0 t

Substitute the equation for v into F_{drag}, which is substituted into the above equation. Next, I set y = 0 and try to solve for t... but the equation is too messy and I can't manage to get just the t's onto one side of the equation.

Am I on the right track? Is there a better way?

Thanks!

 

[cristo]

I've not check the rest of it, but if you want to solve y = \frac{1}{2} (g - \frac{F_{drag}}{m}) t^2 + v_0 t for t with y=0, why not use the quadratic formula?

 

[PhanthomJay]

You seem to be looking only in the vertical direction. Is that the only direction of motion? And you can't use the motion equation for distance that you have noted, since that one was derived for constant acceleration, which you certainly don't have. I'm afraid you'll have to solve the differential equation F_net= mdv/dt in both directions to arrive at a result which involves the hyperbolic functions. Rather difficult at best.

 

[denverdoc]

Not sure I agree. To compute the trajectory would require setting up differential eqns along both axes, and the y would involve a hyperbolic fx. The flight time might be an easier soln, and involve only analysis on the Y axis. In other words the fact that the parabolic trajectory would be foreshortened may have no bearing, just as whether a bullet is fired or simply dropped from the same height of the gun barrel. You may be right, just can't get a good grip on the question.

It would have to be parced into two equations, the ascent where mg is working against and there is a Vy(init) and the descent, where mg and drag are working against each other and Vy(init)=0.

 

[Caspian]

Thanks for all your suggestions. I like denverdoc's suggestion since force and velocity can be broken into x and y components... so force of drag in the x direction won't effect the y direction.

So, here's what I've got for the y direction:
y = \frac{1}{2} (g - \frac{F_{drag}}{m}) t^2 + v_0 t
y = \frac{1}{2} (g - \frac{-B v^2}{m}) t^2 + v_0 t
y = \frac{1}{2} \left(g - \frac{-B \left(\dfrac{1}{3 \dfrac{B}{m} t + C} \right)^{2/3}}{m} \right) t^2 + v_0 t (substituted in v, which was solved for in my previous post. Note that the v and F are just y components)

The only way I can factor out the t is by assuming C = 0. (Is that valid?)
y = \frac{1}{2} g t^2 + \frac{1}{2} \frac{B \left(\dfrac{1}{3 \dfrac{B}{m}} \right)^{2/3}}{m} t^{4/3} + v_0 t

y = \frac{1}{2} g t^2 + \frac{1}{2} \sqrt[3]{B} \frac{\left(\dfrac{1}{3} \right)^{2/3}}{m^{5/3}} t^{4/3} + v_0 t

So, I need to set y = 0 and solve for t (other than the trivial solution of t = 0). I can't use the quadratic formula here... is there another way to solve this expression for t?

denverdoc and PhanthomJay mentioned that this will reduce to an expression which involves a hyperbolic function... but this doesn't look like it would reduce to that form (There's no e's here). Am I doing something wrong?

I really appreciate all your help.

 

[denverdoc]

I suggested such from the following approach:

lets take the second half of the flight after apex has been reached.

mdv/dt=mg-kv(t)^2 where v is in the y component of velocity;

this becomes dv/(g-(k/m)v^2)=dt. let g'=sqrt(g) and c'=sqrt(k/m)

so that dv/[(g'-c'*v(t))*(g'+c'*v(t))]=dt which can be expressed as sum of fractions, and on integration gives rise to ln with numerator and denominator, etc. That should provide explicit function for v(t) in terms of t. For the ascent (the eqn will be different since both mg and kv^2 are negative terms), as we know the limits of integration (Vinit and 0), the time to apex determination should be straightforward. The descent however is a bit trickier, as we do not know the final velocity. But we could use the altitude determined from the ascent phase by integrating once more to solve it I think.