Projectile motion of a basketball shot I'm working on a project which calculates the some statics of a basketball shot. I haven't done physics since high school so I wanted to see if I was on the right track or if I'm completely wrong. Note: this is not a problem for school or nothing like that.
Currently the information I've got to work with is as follows:

*

*Height of the hoop

*Distance from the hoop

*Height of when the ball was released

*Time in the air (can be calculated from when it left the players can till it goes in the hoop)

What I don't have (and trying to find):

*

*Angle of release

*Initial velocity

I was following pretty much whats in this video: https://www.youtube.com/watch?v=fNfkYWqB9w8
but since the basketball hoop is a higher elevation that means I have to find that, correct? Could i just use: $y−y_0=(v_yt)−(\frac{1}{2}gt^2)$ where $y =$ height of basketball hoop and $y_0 =$ height of where ball was released? (then solve for $v_y$)
If so I could just use the remaining formulas of $V_x = \Delta x / \Delta t$ and $a^2 + b^2 = c^2$ to find the angle like in the video.
I guess all I'm really asking is to make sure I'm doing this correctly.
 A: Using the Pythagorean theorem on distances is the wrong idea for finding the angle. With the data you have listed , you have a unique solution for $v_{y0}=\text{your }v_y$ and $v_{x0}= \text{your }V_x$. Then find the magnitude and angle of velocity from the components.
A: $\def\th{\theta}
\def\ra{\rightarrow}$Suppose the ball is thrown from $(0,h)$ to $(d,H)$ under the influence of gravity in time $t$ and that the initial velocity is
${\bf v}_0 = (v_0\cos\th,v_0\sin\th)$.
(In what follows we assume $d>0$ so $-\pi/2<\th<\pi/2$.)
We have
\begin{align*}
d &= v_0 t\cos\th \\
H &= h + v_0 t\sin\th-\frac1 2 g t^2.
\end{align*}
This is a two-by-two nonlinear system of equations for $(v_0,\th)$.
We solve this system with a standard method.
The system is equivalent to
\begin{align*}
v_0 t\cos\th &= d \tag{1}\\
v_0 t\sin\th &= H-h+\frac1 2 g t^2.\tag{2}
\end{align*}
We square each side of (1) and (2), add, and use Pythagoras' theorem with the result
$$v_0^2 t^2 = d^2 + \left(
H-h+\frac1 2 g t^2
\right)^2.$$
Thus,
$$v_0 = \frac{
\sqrt{d^2 + \left(
H-h+\frac1 2 g t^2
\right)}}{t}.$$
If instead we take the ratio of (2) to (1) and solve for $\th$ we find
$$\th = \arctan\frac{H-h+\frac1 2 g t^2}{d}.$$
One can check that these results are dimensionally correct and that they "act right" by taking various limits.
(A) For example, suppose $H=h$ and $t\ra 0$.
We find $(v_0,\th)\ra(d/t,0)$.
If the ball lands at the height from which it was thrown, for short time periods the ball's motion is uniform and horizontal.
(B) If $g\ra 0$ we find
$(v_0,\th)\ra(\sqrt{d^2+(H-h)^2}/t,\arctan((H-h)/d))$.
That is, if there is no gravitational force, the ball's motion is uniform from $(0,h)$ to $(d,H)$.
(C) If $d\ra 0$ (and $H-h+\frac 1 2 g t^2>0$) we find $(v_0,\th)\ra((H-h+\frac 1 2 g t^2)/t,\pi/2)$.
That is, the motion is in the vertical direction and $H=h+v_0 t-\frac 1 2 g t^2$.
This is just one-dimensional kinematics of constant accelerated motion.
