Game Winning Three

The first thing to notice is that we have a choice in both the angle $\theta$ and speed $v$ of launch.

Considering the extreme cases, it is clear that if we shoot the ball at the least possible angle (nearly horizontal) or at the greatest possible angle (just less than vertical), the slightest mistake in angle will lead to the ball deflecting off the rim. Thus, we expect an intermediate launch angle that is least sensitive to errors (diagram below).

For a ball to make it from our hands, a distance $h$ below the rim, to the rim located a distance $d$ away, the following relations must be true

$\begin{aligned} h &= v\sin\theta\ T - \frac12 gT^2 \\ d &= v\cos\theta\ T \end{aligned}$

Solving for $d$ , we find

$d = \frac{v\cos\theta}{g}\left(v\sin\theta+\sqrt{v^2\sin^2\theta + 2gh}\right)$

We want the trajectory $\left(v,\theta\right)$ that's least sensitive to errors in $\theta$ . In other words we need to satisfy the distance relation, and we also need the variation in $d$ with respect to $\theta$ to be a minimum

$\displaystyle\frac{\partial d}{\partial \theta} = 0$

or

$v \cos ^2\theta \left(\frac{v \sin\theta}{\sqrt{2 g \text{h}+v^2 \sin^2\theta}}+1\right)-\sin\theta \left(\sqrt{2 g \text{h}+v^2 \sin ^2\theta}+v \sin\theta\right) = 0$

We can simplify the minimization relation as follows.

Writing $K = \sqrt{2gh + v^2\sin^2\theta}$ , we have

$\begin{aligned} \sin\theta\left(K+v\sin\theta\right) &= v\cos^2\theta\left(\frac{v\sin\theta+K}{K}\right) \\ \sin\theta K &= v\cos^2\theta \\ \sin^2\theta\left(2gh+v^2\sin^2\theta\right) &= v^2\cos^4\theta \\ 2gh\sin^2\theta &= v^2\left(\cos^4\theta - \sin^4\theta\right) \\ 2gh\sin^2\theta &= v^2\left(\cos^2\theta+\sin^2\theta\right)\left(\cos^2\theta-\sin^2\theta\right) \\ 2gh &= v^2\frac{1-2\sin^2\theta}{\sin^2\theta} \\ v &= \sqrt{\frac{2gh}{1-2\sin^2\theta}}\sin\theta \end{aligned}$

Similarly, we can simplify the optimality condition in the vertical direction (i.e. for $\partial h/\partial \theta$ ) to find $v^2 = gd\tan\theta$ . Putting these two relations together, we obtain

$d\tan\theta = 2h\frac{\sin^2\theta}{1-2\sin^2\theta}$

Interestingly, it seems the launch angle will be completely independent of the strength of gravity! The velocity is of course gravity dependent (i.e. through $v^2 = gd\tan\theta$ ), but the launch angle is a purely geometrical consideration.

After some algebra, we find that this equality is solved in the first quadrant by

$\boxed{\theta_\text{safe} = -\tan^{-1}{\dfrac{h^2+\sqrt{h^2\left(d^2+h^2\right)}}{dh}}}$

Using the numbers provided, we find $\theta_\text{safe}\approx 47.98^\circ$ .

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Note, we could also have solved this numerically by simply plotting the contour lines for the distance constraint, and minimization condition in $\left(v, \theta\right)$ space, and looking for their intersection (i.e. the point where both are satisfied).

We do this below, overlaying the numerical result from above as the black dot.

y=xtanθ-gx(1+〖tan〗^2 θ)/(2u^2 ) Now putting x=11,y=1.15 we get; u^2=(11tanθ-1.15)/55(1+〖tan〗^2 θ) Now differentiating u^2 w.r.t θ and equating with 0 for extreme value we get; 11〖tan〗^2 θ-2.3tanθ-11=0 tanθ=1.11 θ=〖48〗^o