A particle is thrown vertically upward. Its velocity at half of the height is 10 m/s. Then the maximum height attained by it : -
`(g=10 m//s^2)`

Options

A. 16m
B. 10m
C. 8m
D. 18m

Answer

b

Explanation

Let a particle gain maximum height H in time t

At `h=(H)/(2),v=10m//s`
Acceleration acting from A to B is -g
`v^(2)=u^(2)+2a(AB)`
`10^(2)=u^(2)-2g((H)/(2))`
For motion A to C
`v^(2)=u^(2)+2a(AC)`
`0=u^(2)-2gH`
From equation ii and i
`100=2gH-2g(H)/(2)`
`gH=100m`
`10m=100m`
H=10m