A ball is projected form the ground at an angle of `45^(@)` with the horizonatl surface .It reaches a maximum height of 120 m and return to fthe ground .upon hitting the ground for the first time it loses half of its kinetic energy immediately after the bounce the velocity of the ball makes an angle of `30^(@)` with the horizontal surface .The maximum height it reaches after the bounce in metres is _______

Answer

`30.00`

Explanation

`H _(1) = ( v ^(2) sin ^(2) 45 ^(@))/( 2 g ) = 120`
` implies ( u ^(2)) /( 4g) = 120. ..(i)`
when half of kinetic energy is lost
`v = (u)/(sqrt2)`
`H _(2) = ((( u)/(sqrt2))^(2) sin ^(2) 30^(@))/( 2 g) = ( u ^(2))/(16g ) ...(ii)`
From (1) and (ii),
`H _(2) = ( H _(1))/( 4) = 30 m`