The equations for the vertical and horizontal components of my initial velocity. Are:

$v_{x,i} = v_i \cos\theta$ and $v_{y,i} = v_i \sin\theta$

And the displacement components are represented as:

$x = x_i + v_{x,i}t + \frac12 a_xt^2$ and $y = y_i + v_{y,i}t + \frac12 a_yt^2$

Ive played around with all these formulas and what not. But Id like to know if there are other formulas that would allow me to find the values of theta that would allow the projectile to reach a certain target if the initial velocity is know. Or perhaps the velocity needed to reach a point, given a set distance and angle (theta).. There is a formula I found that is a variation of the quadratic formula. But it doesnt seem to give good info when the projectile is launched from a higher or lower elevation.

  • 2
    $\begingroup$ Hi, and welcome! Your question will be much easier to read if you use LaTeX for your formulas: $v_x=v\cos(\theta)$ renders as $v_x=v\cos(\theta)$. It would also help if you made it clearer exactly what your question is. $\endgroup$ Sep 29, 2013 at 21:17
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/79012/2451 $\endgroup$
    – Qmechanic
    Sep 29, 2013 at 21:18
  • $\begingroup$ The over question being what are these formulas that give you an angle when all else is know. $\endgroup$ Sep 29, 2013 at 21:44

3 Answers 3


Consider the projectile at an initial position $(x_0, y_0)$, given an initial velocity of $u$ making an angle $\theta$ above the horizontal.

\begin{align} u_x &= u \cos(\theta);\\ u_y &= u \sin(\theta);\\ \end{align}

The velocity remains constant in the $x$ direction, if you neglect dissipative effects like drag.

The velocity in the $y$ direction changes due to gravity:

\begin{align} v_x &= u_x;\\ v_y &= u_y - gt; \end{align}

The x and y displacements can be given as

\begin{align} s_x &= u_x t;\\ s_y &= y_y t - \frac{1}{2}gt^2; \end{align}

The position of the projectile, hence, is:

\begin{align} x &= x_0 + s_x = x_0 + u_x t;\\ y &= y_0 + s_y = y_0 + u_y t - \frac{1}{2}gt^2; \end{align}

Suppose the projectile is launched from a hill 100m above ground level. You want to find the angle of launch which will allow you to hit an object on the ground, 1000m away.

This gives you:

\begin{align} x_0&=0;\\ y_0&=100;\\ x_{final}&=1000;\\ y_{final}&=0; \end{align}

Putting these values in the equations for $x$ and $y$,

\begin{align} 1000 &= 0 + u \cos(\theta) \times t;\\ 0 &= 100 + u \sin(\theta) \times t - \frac{1}{2}gt^2; \end{align}

You now have 2 equations, with 2 variables ($t$ and $\theta$), which you can solve to get the answer. Note: The equation is quadratic in $t$, meaning you'll get 2 values for $t$. One of these can be eliminated (you'll see why when you solve it)


It's almost simpler to write an algorithm an play with it than find new equations. This page for example is a JavaScript cannonball launcher.

Play around with the code to change the cannon's initial height. And then play around with angles until you discover the range you're looking for.


I've personally found the following equation very useful and sometimes invaluable. It is the equation of the trajectory of the projectile:

$$ y = x\tan\alpha - \frac{1}{2}\frac{g x^2}{u^2 \cos^2 \alpha} $$

The origin is assumed to be the point from which the projectile is thrown, $y$ is the upward direction and $x$ points in the direction of $u_x$, with $u$ being the magnitude of the initial velocity with which the projectile is thrown.

  • $\begingroup$ Hmm I am a little confused about the terms you used for your variables.. Please correct me if I'm wrong but I believe you are saying that 'y' is the vertical displacement, 'x' is the magnitude of the horizontal component of the initial velocity vector, 'u' is the magnitude of the actual magnitude of the initial velocity vector, and alpha is the initial angle aka what I am solving for. Correct? $\endgroup$ Oct 2, 2013 at 21:22
  • $\begingroup$ no, 'x' is the horizontal displacement. (x,y) are coordinates of a point on the trajectory of the projectile. $\endgroup$
    – guru
    Oct 2, 2013 at 23:36

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