A projectile is projected and it takes `9 s` to reach in the horizontal plane through the point of projection. In its path it passes a point `P` after `4 s`. The height of `P` above the horizontal plane is

Options

A. `10 m`
B. `50 m`
C. `100 m`
D. `200 m`

Answer

C

Explanation

`O` to `A`: `T = (2u sin theta)/(g) rArr 9 = (2)/(10) u sin theta`
`u sin theta = 45 m//s`
`O` to `P`: `t = 4 s`
`y = u sin theta t - (1)/(2) g t^(2)`
`= 45 xx 4 - (1)/(2) xx 10(4)^(2)`
`= 100 m`