A particle is projected from a stationary trolley. After projection, the trolly moves with velocity `2sqrt(15)` m/s. for an observer on the trolley, the direction of the particle is as shown in the Fig. 5.121 while for the observer on the ground, the ball rises vertically. the maximum height reached by the abll from the trolley is h metre. The value of h will be:

Answer

3

Explanation

`v_(x)=v cos 60^(@)=2sqrt(15)m//s`

`therefore v=4sqrt(15)m//s`
`u_(y)=v sin 60^(@)`
`=4sqrt(15)xxsqrt(3)/(2)=2sqrt(45)m//s`
`H_("max")=(v_(y)^(2))/(2g)=(4xx45)/(20)=9m`