A particle thrown vertically upwards has velocity 10 `ms^(-1)` at half of its height, then maximum height attained by it is :

Options

A. 10 m
B. 20 m
C. 8 m
D. 16 m

Answer

a

Explanation

Let u be the initial velcoity
Max. height,h=`(u^(2))/(2g)`
Applying `v^(2)-u^(2)=2as`
Here v=10 `ms^(-1)` and `S=(h)/(2)`
`100-u^(2)=-2g(h)/(2)`
`100-u^(2)= -3g(h)/(2)`
`100-u^(2)= -g xx(u^(2))/(2g)`
or `u^(2) =200`
`u=sqrt(200)`
`:. h=(200)/(20)=10 m`