A particle is projected vertically from the ground, takes time `t_(1)` upto point A, `t_(2)` from point A to B and time `t_(3)` from point B to highest point and back to the ground. Find the height of the middle point of A and B from the ground.

Options

A. `(g)/(2) [t_(1)^(2) + t_(2)^(2) + 2 (t_(1) t_(2) + t_(2) t_(3) + t_(3) t_(1))]`
B. `(g)/(4) [(t_(1)^(2))/(2) + (t_(2)^(2))/(2) + t_(1) t_(2) + t_(1)T_(3) + t_(2)t_(3)]`
C. `(g)/(2) [t_(1)^(2) + t_(2)^(2) + t_(1)t_(2) + t_(1) t_(3) + t_(2) t_(3)]`
D. none of the above

Answer

B

Explanation

`v_(A) = u - g t_(1)`
`v_(B) = u - g (1_(1) + t_(2))`
`h_(A) = ut_(1) - (1)/(2) g t_(1)^(2)`
`h_(B) = u (t_(1) + t_(2)) - (1)/(2) g (t_(1) + t_(2))^(2)`
`-h_(B) = v_(B) t_(3) - (1)/(2) g t_(3)^(2)`