a 100kg gun fires a ball of 1kg horizontally from a cliff of height 500m. If falls on the ground at a distance of 400m from the bottom of the cliff. The recoil velocity of the gun is (Take g: `10ms^(-2)`

Explanation

Given , mass of the gun `(m_(1))` = 100 kg

Mass of the ball `(m_(2))` = 1 kg
Height of the cliff(h) = 500 m
Horizontal distance travelled by the ball (x) = 400 m
From `" " h = (1)/(2) gt^(2) (because` Initial velocity in downward direction is zero)
`500 = (1)/(2) xx 10t^(2)`
`t = sqrt(100) = 10 s `
From `" " x = ut , u = (x)/(t)= (400)/(10) = 40` m/s
If v is recoil velocity of gun , then according to principal of conservation of linear momentum,
`m_(1) v = m_(2) u`
`v = (m_(2)u)/(m_(1)) = (1)/(100) xx 40 = 0.4` m/s