Can i calculate the initial speed and angle nedeed to reach a specific point with an specific angle? I am trying to build an intelligent ball launcher that can recognize the target and measure its position, so for the control system I want can manipulate the initial velocity and angle, but I  failed in the task to find the form to make this approximations.
The information I have:

*

*$H_{0} =$ initial height

*$\beta$ = target hit angle

*$(x_{f},y_{f})$ = target position

I need to approximate:

*

*$V_{0} =$ initial velocity

*$α =$ throw angle

An solved equation, or a concrete answer that told me why it's impossible do the approximation with the information that I have.
 A: I recommend you begin with the motion of the thrown ball
depending on time $t$. Its position at time $t$ is
$$\begin{align}
x(t) &= V_0 t \cos\alpha \\
y(t) &= H_0 + V_0 t \sin\alpha - \frac{1}{2} g t^2
\end{align}$$
and its velocity at time $t$ is
$$\begin{align}
\dot{x}(t) &= V_0 \cos\alpha \\
\dot{y}(t) &= V_0 \sin\alpha - g t
\end{align}$$
where $g=9.81\text{ m/s}^2$ is the gravity of earth.
Let $t_f$ be the final time when the ball hits the target.
So from $x_f=x(t_f)$ and $y_f=y(t_f)$ you get these two equations:
$$x_f = V_0 t_f \cos\alpha \tag{1}$$
$$y_f = H_0 + V_0 t_f \sin\alpha - \frac{1}{2} g t_f^2 \tag{2}$$
From the velocity components $\dot{x}(t)$ and $\dot{y}(t)$
you get the angle of elevation $\theta(t)$:

$$\tan\theta(t)=\frac{\dot{y}(t)}{\dot{x}(t)}
=\frac{V_0 \sin\alpha - g t}{V_0 \cos\alpha}$$
You want the final angle of elevation to be $\theta(t_f)=\beta$.
So you get the equation
$$\tan\beta=\frac{V_0 \sin\alpha - g t_f}{V_0 \cos\alpha} \tag{3}$$
Now you have done all the physics and arrived at a purely mathematical problem.
You have 3 equations (1, 2, 3) for 3 unknowns ($V_0$, $\alpha$, $t_f$).
So you need to eliminate $t_f$ and resolve for $V_0$ and $\alpha$.
I leave this task to you.
