A bomber plane moving at a horizontal speed of 20 m/s releases a bomb at a height of 80 m above ground as shown in figure. At the same instant, a hunter of negligible height starts running from a point below it, to catch the bomb with speed 10 m/s. After 2s, he realised that he cannot make it, he stops running and immediately holds his gun and fires in such direction so that just before bomb hits the ground, bullet will hit it. What should be the firing speed of bullet? (Take, g = 10 m/s²)

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$10 \sqrt{10} m / s$

$20 \sqrt{10} m / s$

10 m/s

None of these

answer is C.

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Detailed Solution

$\begin{array}{l} T = \sqrt{\frac{2 H}{g}} = \sqrt{\frac{2 (80)}{10}} = 4 s \\ R = uT = 20 \times 4 = 80 m \end{array}$

In 2 s, hunter will travel $10 \times 2 = 20 m$

For bullet, $T = \frac{2 usin θ}{g} = 4 - 2 = 2 s$ …….(i) $\frac{2 usin θ}{g} = 2 s \Rightarrow usin θ = 10 - - - (1)$

$R = \frac{u^{2} \sin 2 θ}{g} = 80 - 20 = 60 m$ ……..(ii)

From Eqs. (i) and (ii), we get

$\frac{\frac{u^{2} 2 \sin θcosθ}{g}}{u \sin θ} = \frac{60}{10}$

$\tan θ = \frac{1}{3} and u = 10 \sqrt{10} m / s$

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