# Finding the Initial Launch Velocity of a Ball Whose Launch Angle is Known and Trajectory Contains a Given Point (Accounting for Air Resistance)

I am working on a project that involves launching a ball through a basket (similar to basketball, but on a smaller scale for now).

I haven't started building the project yet as I wanted to work out all the theory first. However, the system I am thinking of uses a single flywheel configuration, driven by a DC motor, that imparts the launch velocity on a Nerf ball. The launch angle of the ball is controlled by a servo motor. The height of the basket is known, and I am measuring the distance to the basket using a range finder.

For the theory, I wanted to work out a general equation that finds the launch velocity of the ball given the launch angle and X and Y displacement of the basket (Y is the height of the basket and X is the measured distance to the basket). I've already solved this while ignoring air resistance, but since the Nerf balls are very light, I think I should account for air resistance.

I already have the equations for X and Y displacement, accounting for air resistance. They are given below: $$x = v_{x0}\tau(1 - e^\frac{-t}{\tau})$$$$y = (v_{y0} + v_t)\tau(1 - e^\frac{-t}{\tau}) - v_tt$$ Where:

$$v_{x0}$$ = Horizontal component of launch velocity

$$v_{y0}$$ = Vertical component of launch velocity

$$t$$ = Time

$$\tau$$ = Time constant (defined by $$\frac{m}{k}$$ where $$m$$ is the mass of the ball and $$k$$ is a constant defined drag equation (i.e. $$k = \frac{\rho AC_d}{2}$$).

(I got the above equations from this video, and the equations depend on linear air resistance. Please let me know if I should use quadratic air resistance instead).

Now, for the project, I want to figure out a way to use the above equations to calculate the initial launch velocity needed to launch the ball given the X and Y displacement of the basket and the initial launch angle. I rearranged the above equations into the below equations to find the initial components: $$v_{x0} = \frac{x}{\tau(1-e^\frac{-t}{\tau})}$$ $$v_{y0} = \frac{y + v_tt}{\tau(1-e^\frac{-t}{\tau})}-v_t$$

So, can I use the above equations to solve for my initial velocity? I'm a bit worried that the equations don't fully define the parameters of my trajectory in order to solve for the initial launch velocity.

Also, is there a better way to account for air resistance while keeping the math simpler? Again, I was able to figure it out using a regular parabolic trajectory, which was relatively simple math. Are there some tricks that could let me adjust for air resistance without going through complicated math?

First of all, you should write the explicit form of $$v_t$$ in your second equation, $$v_t=v_{y0}e^{-t/\tau}-(1-e^{-t/\tau})g\tau.$$ Now, add this equation to your system: $$\frac{v_{y0}}{v_{x0}}=\tan\theta,$$ where $$\theta$$ is the launch angle. You have 3 equations and 3 unknowns, $$v_{x0}$$, $$v_{y0}$$ and $$t$$ (the time of the impact). You need to solve (probably numerically) for $$v_{x0}$$ and $$v_{y0}$$ and compute the launch velocity, $$v_0=\sqrt{v_{x0}^2+v_{y0}^2}.$$
• Awesome! Thank you! So, does the explicit form of $v_t$ replace the second equation? Also, could I set $t$ as a constant (say, if I want to manually set the time of impact)? And finally, what would be a good approach for solving for $v_x0$ and $v_y0$ numerically? Thanks again for your help! – Brandon Michelsen May 20 '20 at 4:43
• You should replace $v_t$ in the equation, not the full equation. The time $t$ can be computed, i.e. you can't set it, unless you leave $X$ or $Y$ as an unknown instead of $t$. Solving numerically can be done with Newton's method. – Felipe May 20 '20 at 21:23
• So, I would apply Newton's method to solve for $v_{x0}$ and $v_{y0}$ separately? And would I apply the same method to compute $t$, or is there a better method? Also, this does solve for the launch velocity that contains the given point ($x$, $y$), correct? I apologize for all the questions. I haven't been fully introduced to high level physics (this has all been self research at this point). – Brandon Michelsen May 21 '20 at 0:11