A ball in thrown vertically upwards from the ground with some velocity. The velocity of the particle at one fourth of the maximum height is `10sqrt(3) m//s`. Then choose option(s) `(g = 10 m//s^(2))`

Options

A. The maximum height of the particle from the ground is `20m`.
B. The maximum height of the particle from the ground is `10 m`.
C. The velocity with which the ball was thrown is `15 m//s`
D. The velocity with which the ball was thrown is `20 m//s`

Answer

A, D

Explanation

`(A, D)` Let `u` be velocit of throwing and `h` be maximum height.
Then `h = (u^(2))/(2g)`
`rArr V^(2) = u^(2) + 2as`

`(10sqrt(3)^(2)) = u^(2) + 2(-g)"((h)/(4))`
`rArr 300 = u^(2) - 2g." (u^(2))/(2g(4))`
`rArr u^(2) = 400 rArr = u = 20m//s`
`h = 20m`