$5 \sqrt{3}=\frac{(10)^2 \sin 2 \theta}{g} \text { or } \sin 2 \theta=\frac{\sqrt{3}}{2}$
$\therefore 2 \theta=60^{\circ} \text { or } \theta=30^{\circ}$
Two different angles of projection are therefore, $\theta$ and $\left(90^{\circ}-\theta\right)$ or $30^{\circ}$ and $60^{\circ}$.
Two different angles of projection are therefore, $\theta$ and $\left(90^{\circ}-\theta\right)$ or $30^{\circ}$ and $60^{\circ}$.
$T_1=\frac{2 u \sin 30^{\circ}}{g}=1\,s$
$T_2=2 \frac{u \sin 60^{\circ}}{g}=\sqrt{3}\,s$
$\Delta t=T_2-T_1=(\sqrt{3}-1)\,s$
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