For ground to ground projectile, time take by a particle to ge point O to point C is `T_(1)` and during the same motion time taken by the particle to ge from A to B is `T_(2)` then height 'h' is :

Options

A. `(1)/(2)gT_(1)^(2)`
B. `(1)/(4)g(T_(1)^(2)-T_(2)^(2))`
C. `(1)/(8)g(T_(1)^(2)-T_(2)^(2))`
D. `(1)/(8)g(T_(2)-T_(1))^(2)`

Answer

C

Explanation

For ground ……………
`T_(1) = (2V_(y1))/(g)`
`T_(2) = (2V_(y2))/(g)`
`(V_(y2))^(2) = (V_(y1))^(2) - 2gh`