An object is thrown in vertical upward direction with velocity 'u' then the maximum height attained by the object would be....

Options

A. `(u)/(g)`
B. `(u ^(2))/(2g)`
C. `(u ^(2))/(g)`
D. `(u)/(2g)`

Answer

B

Explanation

`2as =v ^(2) -u^(2)`
At maximum height, the final velocity of the object v=0 and for motion in upward direction `a =-g `
`therefore 2 (-g) h _(max) = 0 -u^(2) (because s = h _(max))`
`therefore h _(max) = (u ^(2))/(2g)`