Physics - Kinematics Question with Solution | TestHub

Given,$u=60\mathrm{ m} {\mathrm{s}}^{-1}, R=180 \mathrm{m}$

$\therefore R=\frac{u^2}{g}\sin 2\theta$

$180=\frac{{\left(60\right)}^2}{10}\sin 2\theta$

$\Rightarrow \sin 2\theta =\frac{1}{2}$

$\Rightarrow \theta ={15}^o$

As there are always two directions of projections$\theta$and$\left({90}^o-\theta \right)$for the given range, hence two directions of projection, i.e.${15}^o$and${90}^o-{15}^o,$i.e.${15}^o$and${75}^o.$Let$T_1$and$T_2$be the times of flight in the two cases, where

$T_1=\frac{2u\sin {15}^o}{g}$and$T_2=\frac{2u\sin {75}^o}{g}$

The man is in danger for a time

$T_2-T_1=\frac{2u}{g}\left(\sin {75}^o-\sin {15}^o\right)$

$=\frac{2u}{g}\left(2\cos {45}^o\sin {30}^o\right)$

$\left[\mathrm{u}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{ }\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{t}\mathrm{y},\mathrm{ }\sin C-\sin D=2 \cos \frac{C+D}{2}\sin \frac{C-D}{2}\right]$

$=\frac{2\times 60}{10}\times 2\times \frac{1}{\sqrt{2}}\times \frac{1}{2}$

$=\sqrt{72} \mathrm{s}$