Parabolic projectile equation demonstration question

Question

I was looking at a book of physics and, it will sound dumb, but while I know that the maximum height equation of a projectile is max=(v·senα)/2g, I can't understand how do you get there from 0=v·senα-gt. I need someone to explain it to me step by step.

Answer 1

Suppose you shoot an arrow with initial speed $v$ at the angle $\alpha$ to horizon. We also suppose that there's no interaction with air. We study the second Newton law applied to our arrow: $m\frac{d^2}{dt^2}{\vec x}=m\vec g$ or $ \frac{d^2}{dt^2}{\vec x}= \vec g$.

$\vec x = (h,l)$ - horizontal and vertical coordinate. $\vec g$ has only vertical component , hence we study only the vertical part of our equation:

$\begin{cases}\frac{d^2}{dt^2}h(t) = -g\\ h'(0)=v\sin\alpha\\h(0)=0\end{cases}$

Apparently, $h(t)=tv\sin\alpha-gt^2/2 $. Now it's evident, that the maximum height is when $h'(t)=0$ and $h''(t)<0$ (i.e. the height was increasing before and will decrease after, and right now it doesn't change), which is true for $t_\ast = \frac{v\sin\alpha}{g}$. The height is $h(t_\ast) = \frac{(v\sin\alpha)^2}{2g}$.