# Projectile with air resistance [duplicate]

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.

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

• Question for the student: is $K V_x^2$ the same as $(KV^2)_x$? – dmckee Sep 10 '14 at 2:12
• 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. – dmckee Sep 10 '14 at 3:13