A body is projected vertically upwards at time "t=0" and it is seen at a height H at instants t_(1) and t_(2)seconds during its flight.The maximum height attained is (g is acceleration due to gravity

Options

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

Answer

C

Explanation

`t_(2) = t_(1) + 2t^(.)`
`rArr t^(.) =(t_(2)-t_(1))/2`
Total time taken, to reach point C
`T = t_(1) + t^(.) = t_(1) + (t_(2) - t_(1))/2`
`=(2t_(1) + t_(2) - t_(1))/2 = (t_(1) + t_(2))/2`
Then maximum height attained
`H_("max") =1/2 g(T)^(2) = 1/2 g((t_(1) + t_(2))/2)^(2) = 1/2 g(t_(1) + t_(2))^(2)/4`
`H_("max") = 1/8 g(t_(1) + t_(2))^(2)m`