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So here is one of the questions on my physics exam.

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?

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

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ 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. $\endgroup$ – John Rennie Jun 24 '16 at 18:03
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    $\begingroup$ 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. $\endgroup$ – Anthony64 Jun 24 '16 at 18:07
  • $\begingroup$ What is the direction of air? $\endgroup$ – Anubhav Goel Jun 24 '16 at 18:31
  • $\begingroup$ 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? $\endgroup$ – garyp Jun 24 '16 at 19:13
  • $\begingroup$ What is the air resistance force direction? $\endgroup$ – lucas Jun 24 '16 at 19:49
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I assume that air resistance force is parallel to the velocity vector but in opposite direction.

enter image description here

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 $R=x(t_f)-x_0$.

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  • $\begingroup$ Thank you Lucas. But just to make sure, when you find C (the constant) how do you know that Vy/Vx = ay + g/ax? Also, when you write: ay - CVy + g = 0 and az - CVx = 0, what exactly would I get from solving this? To me there are no unknowns since we know C, Vy and g. I have a feeling that what I'm trying to look for is the force of gravity (-) the air resistance or something similar to this. $\endgroup$ – Anthony64 Jun 25 '16 at 7:08
  • $\begingroup$ Are you familiar with differential equations? Can you solve two equations above? $\endgroup$ – lucas Jun 25 '16 at 8:07
  • $\begingroup$ Differential equations no. I am currently self-teaching them to myself but have started just recently. But doesn't the "or" mean you can use either of the the 2 sets of equations you posted? $\endgroup$ – Anthony64 Jun 25 '16 at 8:25
  • $\begingroup$ Yes, "or" means what you are saying. But, you cannot solve them without any knowledge about differential equations. Sorry, I don't know what we can do for you. The problem isn't at high school level. By the way, I will think more and if I can find an easy method, I will edit my answer. Sorry again. $\endgroup$ – lucas Jun 25 '16 at 9:48
  • $\begingroup$ Thank you very much Lucas. Actually, this question was given to me for one of my exams and I am in 11th grade. If you can find an easier answer, that would be great but if not, don't drive yourself crazy over it! hahaha... One important thing (possibly) is that I am in a school in Barcelona, Spain and the level might be different than what your thinking. $\endgroup$ – Anthony64 Jun 26 '16 at 20:15

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