The maximum ordinate of a point on the graph of the function $f(x)=\sin x(1+\cos x)$, is

$\frac{3 \sqrt{3}}{4}$

We have,

$f(x) =\sin x(1+\cos x)$

$\Rightarrow f(x) =\sin x+\frac{1}{2} \sin 2 x$

$\Rightarrow f'(x)=\cos x+\cos 2 x$ and $f''(x)=-\sin x-2 \sin 2 x$

For local maximum or minimum, we must have

$f'(x)=0$

$\Rightarrow \cos x+\cos 2 x=0$

$\Rightarrow 2 \cos ^2 x+\cos x-1=0$

$\Rightarrow (2 \cos x-1)(\cos x+1)=0$

$\Rightarrow \cos x=\frac{1}{2},-1 \Rightarrow x=\frac{\pi}{3}, \pi, \frac{5 \pi}{3}$

Now,

$f''\left(\frac{\pi}{3}\right)=-\sin \frac{\pi}{3}-2 \sin \frac{2 \pi}{3}<0, f''(\pi)=0$

and, $f''\left(\frac{5 \pi}{3}\right)=-\sin \frac{5 \pi}{3}-2 \sin \frac{10 \pi}{3}>0$

Thus, f(x) attains a local maximum at $x=\frac{\pi}{3}$

∴ Maximum ordinate = $f\left(\frac{\pi}{3}\right)=\sin \frac{\pi}{3}\left(1+\cos \frac{\pi}{3}\right)$

$=\frac{\sqrt{3}}{2} \times \frac{3}{2}=\frac{3 \sqrt{3}}{4}$