Average Landing Distance

Let $y$ be a function of a quantity $x$ . Then, the average value of $y$ as $x$ varies from $x_1$ to $x_2$ is defined as :-

$y_{avg.} = \dfrac{\displaystyle\int_{x_1}^{x_2}f(x)dx}{\displaystyle\int_{x_1}^{x_2}dx}$

= $\dfrac{\displaystyle\int_{x_1}^{x_2}f(x)dx}{{x_2}-{x_1}}$

Over here, $y$ is the Range of the projectile(distance of the landing point from the launch point) , which varies with the angle of projection $\theta$ as :-

$R = \dfrac{v^2\sin2\theta}{g}$

Here, $\theta$ varies from $0$ to $\dfrac{\pi}{2}$

So :-

$Range_{avg.} = \frac{\displaystyle\int_{0}^{\dfrac{\pi}{2}}\dfrac{v^2\sin2\theta}{g}d\theta}{\displaystyle\int_{0}^{\dfrac{\pi}{2}}d\theta}$

= $\dfrac{\dfrac{v^2}{g}\displaystyle\int_{0}^{\dfrac{\pi}{2}}\sin2\theta d\theta}{\dfrac{\pi}{2} - 0}$

= $\dfrac{2v^2}{\pi g}\displaystyle\int_{0}^{\dfrac{\pi}{2}}\sin2\theta d\theta$

This is an ultimately basic integral. On computing it, we get $1$ .

So, the average range becomes $\dfrac{2v^2}{\pi g}$

Comparing with the given expression we find :-

$\alpha = \dfrac{2}{\pi}$

$\alpha = 0.637$