# 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.

• 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. Commented Sep 10, 2014 at 1:49
• Question for the student: is $K V_x^2$ the same as $(KV^2)_x$? Commented Sep 10, 2014 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. Commented Sep 10, 2014 at 3:13