Ytukyg

I'm trying to derive a formula to find the angle $θ$ required to hit a target that may be higher or lower than the initial launch position.

My known variables are $g, v_0, y_0, y_f, x_0, x_f$ where:

• 
  $θ$ is the initial launch angle and the variable being solved for


• 
  $g$ is the gravity constant


• 
  $v_0$ is the initial angular velocity


• 
  $y_0$ is the initial launch elevation


• 
  $y_f$ is the final landing position (in such cases where the projectile lands lower or higher than the launch position)


• 
  $x_0$ is the initial horizontal launch displacement


• 
  $x_f$ is the horizontal landing position, or range


• 
  $t$ is the time in seconds during the path of motion

I started by converting the formula for x position into one for time ($x = x_0 + v_{x0}t$ becomes $t = frac{x}{x_0 + v_{x0}}$

I set the equation for total flight time ($frac{v_{y0}}{g} + frac{sqrt 2h}{g}$) equal to this equation, since $v_{xf} = v_{x0}$ and $x_f$ is known. Leading me to $frac{x}{x_0 + v_{x0}} = frac{v_{y0}}{g} + frac{sqrt 2h}{g}$

I took the equation for maximum height, $frac{gt^2}{2}$ and substituted the expression for time into it, giving me $frac{gx^2}{2x_0^2 + 4x_0v_{x0} + v_{x0}^2}$. Since $v_{y0}$ can be expressed in terms of $v_{x0}$, I also substituted that in ($v_{x0} = v_0cdot cosθ$ and $v_{y0} = v_0cdot sinθ$ so $v_{y0} = v_{x0}cdot tanθ$.

From there I continued to simplify and rearrange until I ended up with $θ = cos^{-1}frac{2gx^2}{v_0}$, however this doesn't give me the correct answer and this is my third attempt over the past several hours.

I've been using the following values in a standard trajectory formula that finds $x_f$ from $θ, v_0, y_0, x_0$:

• $g = 9.81 m/s$


• $θ = 70 degrees$


• $v_0 = 75 m/s$


• $y_0 = 3 m$


• $y_f = 0 m$


• $x_0 = 0 m$

When I plug these into a standard trajectory formula, I find that the range is 369.65957m, which I have been using for $x_f$. I expect to find $θ approx 70$ when I solve for $θ$ using $v_0$ and $x_f$, but other formulas I have found on the web end up giving me ~45 or ~89 degrees.

Thanks for any help!

Move the launch coordinates to $(0,0)$ by setting $x_f=x_f-x_0$, $y_f=y_f-y_0$ and $x_0=y_0=0$.

We can then write the following equations for the projectile's $x$ and $y$ components:

$$y= (v_0sin theta) t- frac{1}{2}gt^2$$
$$x = (v_0cos theta) t$$

Let us see at what time $t_1$ the projectile is at the target $x$-coordinate, i.e when is $x=x_f$:
$$ t_1 = frac{x_f}{v_0 cos theta}$$
Inserting $t_1$ into the equation for the $y$ component of the projectile, must give the height $y_f$ of the projectile at this time:
$$y_f= (v_0sin theta) frac{x_f}{v_0 cos theta}- frac{1}{2}g(frac{x_f}{v_0 cos theta})^2 tag{1}$$
Remembering that $$frac{1}{cos^2 theta}=1+tan^2 theta$$
we can rearrange equation $1$ to become: $$gx_f^2tan^2 theta - 2v_0^2x_f tan theta + 2v_0^2y_f + gx_f^2 = 0$$
Solving this quadratic equation for $tan theta$ gives the formula in the Wiki link I gave in my comment above, i.e. $$theta = arctan left( frac{v_0^2 pm sqrt{v_0^4 - g(gx_f^2+2y_fv_0^2)}}{gx_f} right)$$