Given the probability density of launching a projectile at an angle $\theta$ directed above the horizontal such that $ f(\theta) = \begin{cases} K, & \text{$\frac{\pi}{6}<\theta<\frac{\pi}{3}$} \\ 0, & \text{otherwise} \\ \end{cases} $. The maximum height reached by the projectile is $y=\frac{u^2}{2g}sin^2\theta$ where u is the initial speed and g is the magnitude of the gravitational acceleration.
I am first tasked to determine K and upon solving it, I found out that
$$K=\frac{6}{\pi}$$
For my second task I have to determine the probability density of the maximum height $h(y)$. I know that y is at maximum when $\theta$ is at maximum. So, it should be $\theta=\frac{\pi}{2}$ so that $sin\theta=1$. I want to ask if it's correct to assume that $$h(y) = \begin{cases} \frac{u^2}{2g}sin^2\theta, & \text{$0<\theta<\frac{\pi}{2}$} \\ 0, & \text{otherwise} \\ \end{cases} $$ If not, then do I have to relate the bounds to the constraint given on $f(\theta)$? If my understanding on how to find the probability density of $h(y)$ is wrong then can anyone guide me on how to attack this problem properly? Thank you for reading this post up until this point.
Your value for $K$ is correct.
Hint:
We have $\theta\sim\operatorname{Uniform}(\pi/6,\pi/3)$ and $Y=c \sin^2\theta$ where $c>0$. Then $Y$ is a monotone transformation of $\theta$ and has probability density $$ f_Y(y)=\frac{3}{\pi\sqrt{y(c-y)}},\quad y\in[c/4,3c/4]. $$ You're job is now to figure out why that's the solution. You may want to look up the change of variables formula for probability densities.