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If you have a projectile with these variables. $x_0 = 1v_{0x} = 70, y_0 = 0, v_{0y} = 80, a_x = 0, a_y = -9.8$ I know how to plot these points with this equation. $$ x = x0 + (v_{0x})t + 1/2((a_x)t^2) $$ $$ y = y0 + (v_{0y})t + 1/2((a_y)t^2) $$

I want to add air resistance to this problem and i know its a sphere, so the drag coefficient is 0.47, and lets say the area is 0.5. I use this equation to find the resistance. $$K = 1/2*C_p*A_p$$ where $C_p$ is the drag coefficient and $A_p$ is the area of the sphere. I then try to find the velocity of x and y by using these equations. $$F_dx = KV^2_x$$ $$F_dy = KV^2_y$$ I then plug these in back into my initial x and y equations $$ x = x0 + (v_{0x})t - 1/2((F_dx/m +a_x)t^2) $$ $$ y = y0 + (v_{0y})t - 1/2((F_dy/m +a_y)t^2) $$

I am having a hard time getting the right numbers and pictures when i use these equations. Am i doing something wrong here? Will someone please help me. I would really appreciate any help.

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marked as duplicate by alemi, Brandon Enright, Qmechanic Sep 10 '14 at 5:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The quadratic solution only applies for the problem without air resistance. With linear air resistance you will get projectiles which will slow down exponentially, so you will have to solve the differential equation with resistance first and use those solutions which contain the area and drag coefficient. $\endgroup$ – CuriousOne Sep 10 '14 at 1:49
  • $\begingroup$ Duplicate of: physics.stackexchange.com/q/134147, please don't ask the same question twice, instead edit your old question. $\endgroup$ – alemi Sep 10 '14 at 2:07
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    $\begingroup$ Question for the student: is $K V_x^2$ the same as $(KV^2)_x$? $\endgroup$ – dmckee Sep 10 '14 at 2:12
  • $\begingroup$ yes, that would be the same. $\endgroup$ – user2856118 Sep 10 '14 at 3:08
  • $\begingroup$ Recall that for $V$ pointing at an angle $\theta$ with respect to the horizontal we write $V_x = V \cos\theta$. I cheated a little on the second notation. I mean the x component of a force that is proportional to $V^2$ and directed along (or against) the velocity. That makes the two possibilitites $K V^2 \cos^2 \theta$ and $K V^2 \cos \theta$. Not the same except in a few special cases. $\endgroup$ – dmckee Sep 10 '14 at 3:13
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Those first two equations you mentioned only work in the case of constant acceleration (for more info on this type of kinematics, go here: Can the equations of motion be used for both instantaneous and average quantities?). In your case, we clearly don't have constant acceleration if the force (which defines the acceleration) depends on how fast the object is going. Just picture it this way: first the object starts out with some speed, so there's air resistance which slows it down, so now it has less speed, therefore the air resistance is lesser. So there's a changing acceleration, and you can't apply those seemingly standard kinematics equations. I'm afraid if you don't know a bit of differential equations (or at least basic differential calculus) it'll be impossible for you to understand how to solve the problem (so learn calculus!). If you do know calculus, here's a really nice look at different cases with quadratic air resistance (the type of air resistance that's acting in your problem). Also, as is mentioned in the page I linked to, you'll find that in your particular case which is fairly general, there is no general solution (pardon the redundancy) to the differential equation that comes out of your situation; you can get an approximate numerical solution though.

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