• question_answer A bomber plane is moving horizontally with a speed of $500m/s$ and bomb released from it, strikes the ground in 10 seconds. Angle at which it strikes the ground is $(g=10m/{{s}^{2}})$: A)  ${{\sin }^{-1}}\frac{1}{5}$                     B)  $ta{{n}^{-1}}\frac{1}{5}$C)  $ta{{n}^{-1}}\,5$              D)  $ta{{n}^{-1}}\,1$

Here: Speed of the bomber plane $V=\text{ }500\text{ }m/s$. Time taken by the bomb to strike the ground $=10\sec .$                         $g=10\,m/{{s}^{2}}$ $\therefore$   Time is given by             $t=\sqrt{\frac{2h}{g}}$ $\Rightarrow$               $10=\sqrt{\frac{2h}{g}}$ $\Rightarrow$               $\frac{2h}{g}={{(10)}^{2}}=100$ $\Rightarrow$               $2h=100\times 10$ $\Rightarrow$               $h=500\,m.$ and vertical velocity is given by             $V=\sqrt{2gh}$                 $V=\sqrt{2\times 10\times 500}=100\,m/s.$ Therefore,     $\tan \theta =\frac{vertical\text{ }velocity}{horizontal\text{ }velocity}$                 $=\frac{100}{500}=\frac{1}{5}$                 $\theta ={{\tan }^{-1}}\left( \frac{1}{5} \right)$