Average Landing Distance (Part 2)

Landing distance $d=\frac{{2{v^2}\sin \theta \cos \theta }}{g}$ $\begin{gathered} {d_{av}} = \frac{1}{{\frac{\pi }{2}{v_{\max }}}}\int_0^{\frac{\pi }{2}} {\int_0^{{v_{\max }}} {\frac{{2{v^2}\sin \theta \cos \theta }}{g}dvd\theta } } \\ = \frac{4}{{\pi g{v_{\max }}}}\int_0^{{v_{\max }}} {{v^2}dv\int_0^{\frac{\pi }{2}} {\sin \theta \cos \theta d\theta } } \\ = \frac{{4v_{_{\max }}^2}}{{3\pi g}}\int_0^{\frac{\pi }{2}} {\sin \theta \cos \theta d\theta } \\ let\,u = \sin \theta \\ du = \cos \theta d\theta \\ d\theta = \frac{{du}}{{\cos \theta }} \\ {d_{av}} = \frac{{4v_{\max }^2}}{{3\pi g}}\int_0^1 {udu} = \frac{2}{3}\frac{{v_{_{\max }}^2}}{{\pi g}}\\ \end{gathered}$

First we must consider the fact that the probability of choosing some angle $\theta$ between $0$ and $\frac{\pi}{2}$ is completely uniform. The same goes for choosing some velocity $v$ between $0$ and $v_{max}$ . We calculate a probability density function for choosing a $(\theta, \ v)$ pair by realizing that $P(\theta, \ v) \ = \ C$ , where $C$ is some constant:

$\displaystyle\int_{0}^{v_{max}} \ \displaystyle\int_{0}^{\pi/2} \ C \ \text{d}\theta \ \text{dv} \ = \ C \ \displaystyle\int_{0}^{v_{max}} \ \frac{\pi}{2} \ \text{dv} \ = \ C\frac{\pi \ v_{max}}{2} \ = \ 1 \ \Rightarrow \ C \ = \ \frac{2}{\pi \ v_{max}}$

Notice how I set the integral equal to one, as the probability density function has to be normalized. Next, we have to set up the following equations:

$x \ = \ vt\cos \theta \ \ \ \ \ \ \ y \ = \ vt\sin \theta \ - \ \frac{1}{2}gt^2$

From these equations, we can easily get:

$t \ = \ \frac{x}{t\cos \theta} \ \Rightarrow \ y\ = \ \frac{x}{t\cos \theta} \ v\sin \theta \ - \ \frac{1}{2}g\Big( \frac{x}{t\cos \theta} \Big)^2 \ = \ x\tan \theta \ - \ \frac{gx^2}{2v^2\cos^2\theta}$

Since we are interested in range, we can set $y \ = \ 0$ and create a function of the form $x(v, \ \theta)$ :

$x\tan \theta \ - \ \frac{gx^2}{2v^2\cos^2\theta} \ = \ 0 \ \Rightarrow \ x\tan \theta \ = \ \frac{gx^2}{2v^2\cos^2\theta} \ \Rightarrow \ x \ = \ \frac{2v^2\sin \theta \cos \theta}{g}$

We can now integrate the integral of this form to find the expectation value of our function:

$\langle x \rangle \ = \ \displaystyle\int_{a}^{b} \ x P(x) \ \text{dx} \ = \ \displaystyle\int_{0}^{v} \ \displaystyle\int_{0}^{\pi/2} \ \frac{2}{\pi \ v_{max}} \ \frac{2v^2\sin \theta \cos \theta}{g} \ \text{d}\theta \ \text{dv} \ = \ \frac{2}{\pi \ v_{max}} \ \displaystyle\int_{0}^{v} \ \frac{2v^2}{g} \ \displaystyle\int_{0}^{\pi/2} \ \sin \theta \cos \theta \ \text{d}\theta \ \text{dv}$

Now, using some simple $u$ -substitution :

$u \ = \ \sin \theta \ \ \ \ \ \ \text{du} \ = \ \cos \theta \ \text{dx}$ $\Rightarrow \ \displaystyle\int_{0}^{\pi/2} \ u \ \text{du} \ = \ \frac{u^2}{2} \ \biggr\rvert_{0}^{\pi/2} \ = \ \frac{\sin^2 \theta}{2} \ \biggr\rvert_{0}^{\pi/2} \ = \ \frac{1}{2}$

We can substitute this back into our original equation:

$\Rightarrow \ \frac{2}{\pi \ v_{max}} \ \displaystyle\int_{0}^{v} \ \frac{2v^2}{g} \ \frac{1}{2} \ \text{dv}$

Taking the remaining integral:

$\frac{2}{\pi \ v_{max}} \ \displaystyle\int_{0}^{v} \ \frac{2v^2}{g} \ \frac{1}{2} \ \text{dv} \ = \ \frac{2}{\pi g v_{max}} \ \displaystyle\int_{0}^{v_{max}} \ v^2 \ \text{dv} \ = \ \frac{2}{\pi g v_{max}} \ \frac{v_{max}^3}{3} \ = \ \frac{2v_{max}^2}{3 \pi g}$

Looking the original question, the answer to the problem is given in the form $a \ + \ b \ = \ 2 \ + \ 3 \ = \ 5$ .