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How to derive equations of trajectory of projectile $y=x\tan\theta-\dfrac{gx^{2}}{2 v^{2} \cos ^{2} \theta}$ or $y=x\tan\theta\left(1-\dfrac{x}{R}\right)$, where $R$ is the horizontal range? I am not able to derive any one of the two and the derivations are not given in my textbook. I have searched these equations but I did not find it anywhere including Wikipedia. I know that the trajectory of projectile is a parabola. My question is how to derive $$y=x \tan \theta-\dfrac{g x^{2}}{2 v^{2} \cos ^{2} \theta} \tag{1}$$ $$y=x \tan \theta\left(1-\dfrac{x}{R}\right) \tag{2}$$

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  • $\begingroup$ Do you prefer a link/a picture from a material/complete derivation in mathjax? i don't remember seeing a derivation question in a long time here.... Or you could search "Prove a projectile motion is parabolic in path" $\endgroup$ Oct 3 at 8:12
  • $\begingroup$ @ACB I am sorry, I didn't read the faq tag description, I thought something else. I have edited my question to remove the Faq tag from my question. $\endgroup$ Oct 18 at 19:44
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Whenever we have to find out equation of path of motion, we have to eliminate all parametrical variables, (like time), except for $x,y$ in our kinematical equations.

In projectile motion, we specifically have two equations, one for each direction:

$$x=u(\cos\theta) t$$ $$y=u(\sin\theta)t-\frac{1}{2}gt^2$$

Now as i mentioned, we must think of eliminating our variable of time $t$.

Doing it we will get: $$y=x \tan \theta-\dfrac{g x^{2}}{2 v^{2} \cos ^{2} \theta} $$ Now, we know: $$R=\frac{u^2\sin(2\theta)}{g}$$

Adjusting in our equation of path, with some algebrae:

$$y=x \tan \theta\left(1-\dfrac{x}{R}\right)$$

Hope you got the idea!

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Hints:

For 1) set up equations from the equations of motion for the distance travelled in the x direction in terms of time. Same for the y direction. Then make time the subject of the x direction formula and substitute into the y.

For 2) substitute the formula for range into 2) and see if you can get to 1), then reverse the steps if you want 1) to 2).

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