I'm having problems to proof the equation for maximum height which is given as follows:
$$H_{\max}=\frac{v_o\sin^2\omega}{2\times g}$$
starting from here (which is the equation for $y$):
$$y=v_{oy}t-\frac{1}{2}gt^2$$
I am confused whether if the speed on $y$-axis becomes $0$ in the maximum height but would not this cancel the first term in the above equation?.
What would be the way to go?
Edit:
Although there can be different ways to assess this situation. What would be the one that explains what is happening to the vectors while the projectile is in the top?
Best Answer
The vertex of the parabola occurs at $t=-\dfrac{b}{2a} =\dfrac{v_{0y}}{g}$. Substituting $v_{0y}=v_0\sin\omega$ and evaluating the function there should give you what you're after.