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

Here , `m_(1) = 100 kg , m_(2) = 1 kg , h = 500 m ,`
`x = 400 m`
Let u be velocity of ball and t be time taken by the ball to hit the ground
From `h = (1)/(2) g t^(2)`
`500 = (1)/(2) x 10t^(2)`
`t = sqrt(100) = 10 s`
From `x = ut, u = (x)/(t) = (400)/(10) = 40 m//s`
If is recoil velocity of gun then according to principle of conservation of linear momentum `m_(1) upsilon = m_(2) u`
`upsilon = (m_(2)u)/(m_(1)) = (1)/(100) xx 40 = 4.0 m//s`
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