homework and exercises - Projectile motion formulas

Assuming:

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.

Answer

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)