Time of flight with air resistance

In summary, you need to compute the time of flight of a projectile which is subject to air resistance. You started by noting that F_{drag} = -B v^2 and a = \frac{\partial v}{\partial t}. You then integrated and solved for v. You found that v = \sqrt[3]{\dfrac{1}{3 \dfrac{B}{m} t + C}} and that you can solve for y by substitution. You found that y = \frac{1}{2} (g - \frac{F_{drag}}{m}) t^2 + v_0 t and that y = \frac{1}{2} (g - \frac
  • #1
Caspian
15
0
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:

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

integrate and solve for v...

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

I then plug this into the kinematics equation for position:

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

Substitute the equation for v into [tex]F_{drag}[/tex], 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!
 
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  • #2
I've not check the rest of it, but if you want to solve [tex]y = \frac{1}{2} (g - \frac{F_{drag}}{m}) t^2 + v_0 t[/tex] for t with y=0, why not use the quadratic formula?
 
  • #3
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.
 
  • #4
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.
 
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  • #5
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:
[tex]y = \frac{1}{2} (g - \frac{F_{drag}}{m}) t^2 + v_0 t[/tex]
[tex]y = \frac{1}{2} (g - \frac{-B v^2}{m}) t^2 + v_0 t[/tex]
[tex]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[/tex] (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?)
[tex]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[/tex]

[tex]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[/tex]

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.
 
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  • #6
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.
 

1. What is the Time of Flight with Air Resistance?

The Time of Flight with Air Resistance is the amount of time it takes for an object to travel through the air and reach its final destination while experiencing the force of air resistance. This is typically longer than the Time of Flight without air resistance due to the additional force acting on the object.

2. How does air resistance affect the Time of Flight?

Air resistance is a force that opposes the motion of an object through the air. This force increases as the speed of the object increases. As a result, the Time of Flight with air resistance will be longer than the Time of Flight without air resistance because the force of air resistance slows down the object.

3. What factors affect the Time of Flight with Air Resistance?

The Time of Flight with air resistance is affected by several factors, including the surface area of the object, the density of the air, and the velocity of the object. Objects with larger surface areas or moving through denser air will experience more air resistance and therefore have a longer Time of Flight.

4. How can the Time of Flight with Air Resistance be calculated?

The Time of Flight with air resistance can be calculated using the equation t = (2v0sinθ) / g + (2v0cosθ) / k, where t is the time of flight, v0 is the initial velocity, θ is the launch angle, g is the acceleration due to gravity, and k is the air resistance coefficient. It is important to note that this equation is an approximation and may not be entirely accurate in all situations.

5. How can the Time of Flight with Air Resistance be minimized?

The Time of Flight with air resistance can be minimized by reducing the effects of air resistance. This can be achieved by using a more aerodynamic object, increasing the initial velocity, or reducing the launch angle. Additionally, launching the object in a vacuum or low air density environment can also minimize the effects of air resistance and reduce the Time of Flight.

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