Question

A bomber plane moving at a horizontal speed of 20m/s releases a bomb at a height of 80 m above ground. At the same instant a hunter of negligible height starts running from a point below it to catch the bomb with a speed of 10m/s. After two seconds he realizes that he cannot mak 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 the bullet? Take g = 10m/s^2

Answer 1

When bomb is dropped from 80 m height the total time taken by the bomb to reach the ground is

$h = \frac{1}{2}gt^2$

$80 = \frac{1}{2}10t^2$

$t = 4 s$

Total range covered by the bomb in horizontal direction is given as

$R = v_x * t$

$R = 20 * 4 = 80 m$

now the distance covered by the man in 2 s with speed 10 m/s is given as

$d = v* t = 10 * 2 = 20 m$

so the total distance that bullet should cover in next 2 s is 80 - 20 = 60 m

so bullet should cover 60 meter distance in 2 s

so speed of bullet is given as

$v = \frac{d}{t}$

$v = \frac{60}{2} = 30 m/s$

so the speed of bullet will be 30 m/s

Answer 2

Answer:

Explanation

initial speed of bomb=20m/s

height=80 m

Therefore time of flight:-

80=0*t+1/2*g*t^2

t=4 s

distance travelled by hunter in 2s=10*2=20m

Range of bomb=20*4= 80m

Therefore distance left to cover is 60m

So V horizontal=60/2=30m/s

V vertical:-

Applying S=ut+1/2at^2

0=v*2-1/2*g*4

v=10m/s

Therefore resultant velocity=(Vx^2+Vy^2)^1/2

=( 30^2+10^2)^1/2

=(900+100)^1/2

=(1000)^1/2

=10√10 m/s

Thank you