Collision ball with backboard I am doing a physical study to study the movement of basketball.
When it hits the backboard, its velocity and angle of reflection change.
How can I calculate the new velocity and angle of the ball after hitting it with an immobile object.
Knowing that the study will be two-dimensional or three-dimensional.

 A: First you can assume the balls don't rotate. This is not true in general but this makes the math much easier.
First step is to decompose the velocity at the time of collision into two components. A component normal to the wall $v_\perp$ and a component parallel to the wall $v_{\parallel}$. Like follows:

In case the ball doesn't rotate we have $v_\parallel'=v_\parallel$ where the prime indicates the velocity after the bounce. The normal component will be reflected and it will generally be slightly smaller if energy is lost in other words $$v_\perp'=-e\cdot v_\perp$$
where $e$ is the coefficient of restitution which is always $0\leq e\leq 1$. When $e=1$ we have an elastic collision and no energy is lost. When $e=0$ all energy is lost and the ball drops straight down like it is made out of clay. To determine $e$ you would generally have to do some experiment or cite it from literature.
Background
To know why the normal component reflects we can look at a more general collision. Let's consider two balls with masses $m_1,m_2$ and velocities $v_1,v_2$ which become $v_1',v_2'$ after the collision. During the collision there is a normal force between the balls. So any change in velocity must be along the normal direction (if the balls are rotating there can be force in the parallel direction as well). So again we decompose the velocities in a normal and tangential direction. The tangential velocities stay the same. If we assume all energy is conserved ($e=1$) in combination with momentum conservation we can derive the following formula for the perpendicular components:
\begin{align}
v_1'=\frac{m_1-m_2}{m_1+m_2}v_1+\frac{2m_2}{m_1+m_2}v_2\\
v_2'=\frac{2m_1}{m_1+m_2}v_1+\frac{m_2-m_1}{m_1+m_2}v_2
\end{align}
Here you should read $v_1$ and $v_2$ as $v_{\perp\ 1}$ and $v_{\perp\ 2}$ but this becomes too messy to write out. A collision with a wall is basically a collision with a very heavy object because the wall is attached to the earth so it is basically colliding with the entire earth. If $m_2$ becomes very large you get
$$\frac{m_1-m_2}{m_1+m_2}\approx-1,\quad\frac{2m_1}{m_1+m_2}\approx 0$$
Try this on a calculator to convince yourself. Because $v_2=0$ the equations become the following
\begin{align}
v_1'&=-v_1\\
v_2'&=0
\end{align}
A: The collision of a ball at an angle with a flat surface is pretty much impossible to predict.  It is subject to two impulses; one from the normal force which will reverse the normal component of velocity (but the magnitude will probably decrease), and the other from friction (parallel with the surface). The magnitude and direction of the friction force will depend on the initial velocity and spin of the ball, and it will cause a change in the spin as well as a change in the parallel component of velocity.
