A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of 400 m from bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity = 10 `ms^(-2)` )

Explanation

Mass of gun `m_(1) =100 kg`
mass of ball `m_(2) =1 kg`
Height of the cliff h =500 m
x = 400m
Initially vertical component of velocity is zero (in downard direction )
`h= (1)/(2) gt^(2)`
`500 = 5t^(2)`
`t= sqrt(100) = 10 s`
For horizontal motion of ball
x= ut
`u = (x )/( t ) = (400)/(10) =40 m//s`
Let v is recoil velocity of gun
By law 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`