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

[30m]

Explanation

[30 m]
Let particle be projected with speed u at an angle `45^(@)` with the horizontal . Maximum height attained is given to be 120m .
`H=(u^(2) sin ^(2) 45^(@))/(2g)=120m`
`rArr (u^(2))/(4g)=120`
Let `u_(x) and u_(y)` be the two components of velocity of particle after it rebounds from the ground first time . It is given that net velocity makes an angle `30^(@)` with the horizontal .
`rArr (u_(y))/(u_(x))=tan 30^(@) =(1)/(sqrt3)`
`rArr sqrt3u_(y)=u_(x)`
It is given that particle loses half of its kinetic energy due to collision . Hence final K.E. is half of the initial .
`(1)/(2)m(u_(x)^(2)+u_(y)^(2))=(1)/(2) mu^(2)`
Using equation (ii) , we get the following:
`(3u_(y)^(2)+u_(y)^(2))=(u^(2))/(2)`
`u_(y)^(2)=(u^(2))/(8)`
Maximum height attained after hitting the ground can be written as follows:
`H.=(u_(y)^(2))/(2g)`
Using equating (iii), we get
`H.=(u^(2))/(16g)`
New using equation (i) , we get the following:
`H.=(u^(2))/(16g)=(120xx4g)/(16g)=30m`