# 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.

• Where are you getting your data from? Is it a video, by any chance? – Philip Jun 29 '20 at 20:17
• Yup, I'm just recording a video and trying to get some information from it like the launch angle. – iG Cloud Jun 29 '20 at 20:20
• You should definitely check out Tracker Video Analysis if you haven't already! It's a fantastic tool that should do everything you want, especially if you're the one recording the videos. – Philip Jun 29 '20 at 20:22
• Will do, I'm actually trying to create something like that for a learning experience, haha :) – iG Cloud Jun 29 '20 at 20:24
• There are probably two solutions for your problem: one where the basketball is shot at a "shallow" angle, and will probably hit the rim, and one where the basketball is shot at a "steep" angle, and will probably "swish" through the hoop. Which solution are you looking for? – David White Jun 30 '20 at 1:02

$$\def\th{\theta} \def\ra{\rightarrow}$$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.
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.