Problem:
Suppose a projectile is fired at an angle $θ$ from the horizontal with a velocity $v$. The horizontal distance that the projectile travels, $R$, is given by
$$R = \frac{v^2}{g} \sin 2 \theta$$
where $g$ is the acceleration due to gravity ($9.8$ m/s2)
If $ \theta$ is uniformly distributed over the range $(0,π/4)$, find the probability density function of $R$.
. . . . . . . . . . . . . . .
I know the answer involves the change of variable method. I get as far as solving the Jacobian derived to be $\dfrac{2v^2\cos2\theta}{g}$. At this point I am unsure how to proceed further to achieve the answer which is given as $\dfrac{2g}{\pi\sqrt{v4-g^2r^2}}$ for $0 < r <v^2/g$
Any help would be appreciated.
Since $0<\theta < \pi/4$, this means that $0< 2\theta< \pi/2$, and the $\sin$ function is monotonic increasingly in this interval, you can then use the fact that if $p_X(x)$ represents the PDF of a random variable $X$ and $Y = f(X)$ is another random variable, then the PDF of $Y$ satisfies
$$ p_X(x)dx = p_Y(y)dy ~~~\Rightarrow p_Y(y) = p_X(x)\left|\frac{dx}{dy}\right| $$
where the absolute value is there to ensure that $p_Y$ is well defined. In your case we have $x = \theta$ and $y = R = v^2 g^{-1}\sin 2\theta$, therefore
$$ \frac{gR}{v^2} = \sin 2\theta ~~~\Rightarrow~~~ \theta(R) = \frac{1}{2}\sin^{-1}\left(\frac{gR}{v^2}\right) $$
so that the derivative is
$$ \frac{d\theta(R)}{dR} = \frac{g}{2v^2}\frac{1}{\sqrt{1 - g^2R^2/v^4}} = \frac{g}{2\sqrt{v^4 - g^2R^2}} ~~~\mbox{for}~~~ 0 < R < v^2/g $$
Finally, since $\theta$ is uniformly distributed, we have $p_\Theta(\theta) = 1/(\pi/4) = 4/\pi$ for $0<\theta< \pi/4$, and $0$ otherwise. Putting eveything together
$$ p(R) = \frac{2g}{\pi\sqrt{v^4 - g^2R^2}} ~~~\mbox{for}~~~ 0 < R < v^2/g $$