Question

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. 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 two seconds he realized 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 $g=10m/s^2$

• A. $70m/s$
• B. $20\sqrt{10}m/s$
• C. $10\sqrt{10}m/s$
• D. None of these

Answer 1

The correct option is A $70m/s$

Here in this case let the height be H.

$H=\left)U_{0y}\right)t+\frac{1}{2}g{t}^2$, where $U_{0y}=0$m/s

$\Rightarrow H=\frac{1}{2}g{t}^2$

$\Rightarrow 80=5t^2$

$\Rightarrow t=4$ seconds

Let the horizontal displacement of the bomb is X.

So at $t=4s$,

$X=\left(U_{0x}\right)t+\frac{1}{2}g{t}^2$

$\Rightarrow X=(20\times 4)+(5\times 16)$

$\Rightarrow X=160m$.

Let the position of solider after t second is x.

$x=ut$ where, $u=10$ m/s

Hence after $2$ seconds, $x=20$ which implies he will stop running after $20$m. So he has to shoot the bomb within another $2$ seconds at the distance $160-20=140m$

Thus using the speed formula.