Projectile motion - differentiating the equation of trajectory to find the maximum height

Question

I have the equation of trajectory:
$ y = x\tan \theta - {\displaystyle gx^2 \over \displaystyle2u^2\cos^2 \theta}$

I also know that the maximum height is given by:
${\displaystyle u^2 \over\displaystyle 2g \ }\sin^2 \theta $

but I can't figure out how you get that by differentiating the first equation. I know there are different forms of the equation of trajetory, so maybe I should be using a different one? Any help would be much appreciated!

Answer 1

So an straight forward way would be to first find the maximum by solving $$ 0 = \frac{dy}{dx} = \tan{\theta} - \frac{g}{u^2 \cos^2{\theta}} x $$ for $x$ and then plugging this value back into the height relation $y(x)$.