The problem I am trying to solve is to find the velocity needed to hit an elevated target a known distance away from the initial position of a launcher.
We know the height of and distance to the elevated target. We also know the initial launch angle of the projectile. We do not know how long it will take for the projectile to reach the target. We are trying to solve for the initial velocity.
I have scoured the internet (including this stackexchange) for a solution, but all of the solutions I have come across seem to only work when the target is at the same elevation as the launcher (which would not work in my situation).
I have also tried to derive my own equation for the initial velocity. I got pretty close, but it is in a form where I cannot isolate the initial velocity in the equation.
$$v_i^2-v_i^2sin(\theta)-\frac{d^2g^2}{v_i^2cos^2(\theta)}=2gh$$
d is the distance to the target
h is the height of the target
theta is the launch angle
vi is the initial velocity
To clarify, I derived this equation using conservation of energy, and plugging in other equations I derived, but there are many ways to come to this equation.
Thanks!
Best Answer
You already mentioned there are some answers here regarding projectile motion. You cannot expect to find an answer that will exactly match you problem every time. You should try to understand the underlying principle instead.
The easiest way to solve this is to start from equations for projectile motion (pretty standard thing) and then plug $d$ and $h$ in proper places.
Assuming there is no drag and mass does not change, the equations of motion are:
$$x(t) = v_{0,x}t \quad \text{and} \quad y(t) = -\frac{1}{2}gt^2 + v_{0,y}t$$
where
$$v_{0,x} = v_0 \cos\theta \quad \text{and} \quad v_{0,y} = v_0 \sin\theta$$
where $v_0$ is the initial velocity (unknown) and $\theta$ is the initial angle (known).
Now plug $d$ as $x$ and $h$ as $y$:
$$v_0 \cos\theta \cdot t = d \quad \text{and} \quad -\frac{1}{2} g t^2 + v_0 \sin\theta \cdot t = h$$
the solution for $v_0$ is (almost) obvious. From the former equation you get the expression for time
$$t = \frac{d}{v_0 \cos\theta}$$
which you use in the latter equation. The expression for $v_0$ almost immediately follows:
$$\boxed{v_0 = \frac{d}{\cos\theta} \sqrt{\frac{g}{2}\frac{1}{d \tan\theta - h}}}$$
It should be noted that $h$ in the above equation can also be negative.