Projectile motion with air friction force/resistance 
We have to find the x (the distance, if you didn't know that then I'm not sure if you should be doing this problem) that the projectile travel during the time in the air until the time it hits the ground. I can do this no problem without air resistance, but I have no idea what to do if there is. My teacher hinted to me to do it using energy formulas but even after doing that, I got stumped. Can someone please tell me or show me how to do this problem using either motion formulas or energy formulas?
 A: I assume that air resistance force is parallel to the velocity vector but in opposite direction.

We have:
$$a_x=\large{\frac{10\cos\theta}m}$$
$$a_y=\large{\frac{10\sin\theta}m}-g\;\Longrightarrow\;a_y+g=\large{\frac{10\sin\theta}m}$$
$$\tan\theta=\large{\frac{v_y}{v_x}}$$
Then,
$$\large{\frac{v_y}{v_x}}=\large{\frac{a_y+g}{a_x}}\;\Longrightarrow\;\large{\frac{a_y+g}{v_y}}=\large{\frac{a_x}{v_x}}=C\;\textrm{(constant)}$$
Left side of equation above is a function of $y$ and the right side is a function of $x$. So, it must be equal to a constant like $C$. Hence, we will have:
$$a_y-Cv_y+g=0$$
$$a_x-Cv_x=0$$
Or
$$\large{\frac{\mathrm d^2y}{\mathrm dt^2}}-C\large{\frac{\mathrm dy}{\mathrm dt}}+g=0$$
$$\large{\frac{\mathrm d^2x}{\mathrm dt^2}}-C\large{\frac{\mathrm dx}{\mathrm dt}}=0$$
You should solve these differential equations by considering to the initial conditions. Then, you can find the time of falling by substitution $y=y_0-1000$ say $t_f$. Finally, the range that you want is determined $R=x(t_f)-x_0$.
