A particle is thrown at an angle with horizontal from the ground. After `4 s` it reaches to a point `P` and after `5` more seconds it strikes the ground. Find the height of `P` the maximum height attained by the particle.

Explanation

Time taken by the particle to move from `O` to `P` is `4 s` and to `A` is `5 s`, hence `O` to `A` is `9 s`.
Time of flight `= (2u sin theta)/(g) = 9`
`rArr (2)/(10)u sin theta = 9 rArr u sin theta = 45`
`O` to `P` :
`y = u sin theta t - (1)/(2) g t^(2)`
`= 45 xx 4 - (1)/(2) xx 10 xx (4)^(2)`
`= 180 - 80 = 100 m`
`H_(max) = (u^(2) sin^(2) theta)/(2 g) = ((45)^(2))/(20) = 101.25m`