Projectile motion with air friction force/resistance [closed] 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?

closed as off-topic by John Rennie, knzhou, CuriousOne, Gert, ACuriousMind♦Jun 25 '16 at 10:14

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• Hi Anthony and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. – John Rennie Jun 24 '16 at 18:03
• This is not homework. My school year has ended and I am trying to learn more physics over the summer. This is a question from an exam taken 2 months ago and I finally got around to bringing it online. It also isn't a copy, it has the exact same concept, just different values. – Anthony64 Jun 24 '16 at 18:07
• What is the direction of air? – Anubhav Goel Jun 24 '16 at 18:31
• You need some model (equation) for the force of air resistance on your object. Without that you can't make progress. Do you have such a model? – garyp Jun 24 '16 at 19:13
• What is the air resistance force direction? – lucas Jun 24 '16 at 19:49 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 $R=x(t_f)-x_0$.