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.

  • $\begingroup$ Where are you getting your data from? Is it a video, by any chance? $\endgroup$ – Philip Jun 29 '20 at 20:17
  • $\begingroup$ Yup, I'm just recording a video and trying to get some information from it like the launch angle. $\endgroup$ – iG Cloud Jun 29 '20 at 20:20
  • $\begingroup$ 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. $\endgroup$ – Philip Jun 29 '20 at 20:22
  • $\begingroup$ Will do, I'm actually trying to create something like that for a learning experience, haha :) $\endgroup$ – iG Cloud Jun 29 '20 at 20:24
  • $\begingroup$ 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? $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.