A particle is projected from the ground with velocity $u$ at an angle $\theta$ with horizontal. The horizontal range, maximum height and time of flight are $R,H$ and $T$ respectively. They are given by, $R=\frac{u^2\sin 2\theta }{g},H=\frac{u^2{\sin }^2\theta }{2g}\text{and}T=\frac{2u\sin \theta }{g}$
Now keeping $u$ as fixed, $\theta$ is varied from ${30}^{\circ }$ to ${60}^{\circ }$. Then

The correct option is B $R$ will first increase then decrease while $H$ and $T$ both will increase
From ${30}^{\circ }$ to ${60}^{\circ },\sin \theta$ increases
$\therefore T=\frac{2u\sin \theta }{g}$
${\sin }^2\theta$ will also increase from ${30}^{\circ }$ to ${60}^{\circ }$
$\therefore H=\frac{u^2{\sin }^2\theta }{2g}$ will also increase
As ${30}^{\circ }<\theta <{60}^{\circ }\Rightarrow {60}^{\circ }<2\theta <{120}^{\circ }.$
So, $\sin 2\theta$ will increase from ${60}^{\circ }$ to ${90}^{\circ }$ but decreases from ${90}^{\circ }$ to ${120}^{\circ }$
$\therefore R=\frac{u^2\sin 2\theta }{g}$ will first increase and then decrease.