Launching a Projectile with Air Resistance

  • #1
xtraboi
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Hello Physics Forums members,

I am a student in AP Physics C and I was just working out the range of a projectile when air resistance is non-negligible. As of right now I'm going to use the linear model of air resistance to simplify calculations (F_air = -bv).

When drawing the Free Body Diagram for the projectile, I know that the force on the object due to air resistance opposes the direction of motion, so does that mean there would be one Free Body Diagram when the projectile is going up and another when the projectile is going down?

If that's the case, how would I go about finding the total range of the projectile?

My approach is to use the kinematic equation v_f^2 = v_i^2 + 2*a*d twice (one for when the object is going up, and one for when the project is coming down) and substitute the acceleration as calculated from the Free Body Diagram, but I'm not sure if that's the right approach as the force due to air resistance is dependent on the instantaneous velocity at a point. I'm thinking to maybe take the average value of the force due to air resistance, but I'm not even sure what integral would achieve that as we haven't covered what I'm doing in class.

Thanks in advance.
 

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  • #2
PeroK
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In your case, acceleration is not constant, but varies with velocity. The SUVAT equations don't apply and you'll need to set up and solve a differential equation or two.
 
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  • #3
xtraboi
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In your case, acceleration is not constant, but varies with velocity. The SUVAT equations don't apply and you'll need to set up and solve a differential equation or two.
Oh that makes sense. Does that mean I would take the integral of the acceleration found from the free body diagram to get v(t), and then take the integral of that to find x(t)?
 
  • #4
PeroK
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Oh that makes sense. Does that mean I would take the integral of the acceleration found from the free body diagram to get v(t), and then take the integral of that to find x(t)?
Sort of: it means you generate a differential equation and integrate that (twice). That's easier said than done, however.
 
  • #5
Orodruin
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Oh that makes sense. Does that mean I would take the integral of the acceleration found from the free body diagram to get v(t), and then take the integral of that to find x(t)?
Yes
 
  • #6
xtraboi
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Yes
Got it, one more question - doesn't the direction of the force due to air resistance change after the projectile reaches its maximum height? As the projectile is ascending and descending, the force due to air resistance changes direction as it opposes the direction of motion.

If so, how would I account for that when solving for v(t) and x(t)?
 
  • #7
Orodruin
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That's easier said than done, however.
##m\dot v = - bv - mg## seems perfectly solvable to me … the real issues start if you consider a non-linear model for air resistance…
 
  • #8
Orodruin
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Got it, one more question - doesn't the direction of the force due to air resistance change after the projectile reaches its maximum height? As the projectile is ascending and descending, the force due to air resistance changes direction as it opposes the direction of motion.

If so, how would I account for that when solving for v(t) and x(t)?
Yes, it changes direction when v changes sign. This is already built into the term -bv, which also changes direction when v changes sign.
 
  • #9
xtraboi
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Yes, it changes direction when v changes sign. This is already built into the term -bv, which also changes direction when v changes sign.
That makes so much more sense. I never really understood the negative sign on -bv until now, so thank you so much for clearing that up, and thank you to PeroK for offering a valuable insight onto the problem :smile:
 
  • #10
kuruman
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##m\dot v = - bv - mg## seems perfectly solvable to me … the real issues start if you consider a non-linear model for air resistance…
I think there would be a complication in two dimensions. The horizontal and vertical components of the acceleration would have to be proportional to ##\sqrt{v_x^2+v_y^2}## which yields a system of two coupled diferential equations.
 
  • #11
xtraboi
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I think there would be a complication in two dimensions. The horizontal and vertical components of the acceleration would have to be proportional to ##\sqrt{v_x^2+v_y^2}## which yields a system of two coupled diferential equations.
Couldn't you just resolve the horizontal velocity as ##v\cos(\theta)## and the vertical velocity as ##v\sin(\theta)##, and solve the differential equation for both to find x(t) and y(t)? If you were to resolve the forces as well, the ##F_{air}## in the vertical direction would be ##F_{air}\sin(\theta)##, meaning the net force in the vertical direction would be $$\Sigma{F_y} = -mg-bv\sin(\theta)$$
 
  • #12
PeroK
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Got it, one more question - doesn't the direction of the force due to air resistance change ...
I would say the direction of air resistance changes continuously to oppose the continuously changing direction of the velocity.
 
  • #13
Orodruin
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I think there would be a complication in two dimensions. The horizontal and vertical components of the acceleration would have to be proportional to ##\sqrt{v_x^2+v_y^2}## which yields a system of two coupled diferential equations.
As long as you are in the linear regime, this is not the case. This is one of the reasons you run into issues when air resistance becomes non-linear.

Edit: To expand on that. The vector form of the linear air resistance is ##\vec F = -b\vec v##. From this it is pretty clear that the components separate. All components appearing in the speed is exactly offset by the projection onto the relevant component.
 
  • #14
kuruman
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As long as you are in the linear regime, this is not the case. This is one of the reasons you run into issues when air resistance becomes non-linear.

Edit: To expand on that. The vector form of the linear air resistance is ##\vec F = -b\vec v##. From this it is pretty clear that the components separate. All components appearing in the speed is exactly offset by the projection onto the relevant component.
Yes, I agree. I fired off the post in a hurry only to realize that later.
 

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