The directioin of projection of particle is shown in the figure for an observer on trolley. An observer on the grond sees the ball rising vertically. The maximum height reached by the ball as seen from the troelley is

Answer

9

Explanation

`v_x = v cos 60^@ = 2 sqrt(15) ms^(-1)`
`:. V = 4 sqrt(15) ms^(-1)`
`:. V_y = v sin 60^@ = 4 sqrt(15) (sqrt3)/2 = 2 sqrt(45) ms^(-1)`
`H_(max) = v_(y)^(2)/(2g) = (4 xx 45)/20 = 9m`.