A particle is thrown such that for two angles of projection range `R` is same. If `T_(1)` and `T_(2)` are the time of flights and `H_(1)` and `H_(2)` are the maximum hrights corresponding to these angles, find (a) `(T_(1)T_(2))/(R )` and (b) `(4sqrt(H_(1)H_(2)))/(R )`.

Explanation

Let angle of projection be `theta`.
`T_(1) = (2 u sin theta)/(g), H_(1) = (u^(2) sin^(2) theta)/(2g)`
`R = (u^(2))/(g)(2 sin theta cos theta)`
For the same range `R`, another angle of projection will `(90 - theta)`.
`T_(2) = (2 u sin(90 - theta))/(g) = (2 u cos theta)/(g)`
`H_(2) = (u^(2) sin^(2) (90 - theta))/(2g) = (u^(2) cos^(2) theta)/(2g)`
(a) `(T_(1)T_(2))/(R ) = ((2 u sin theta)/(g) xx (2u cos theta)/(g))/((u^(2))/(g) (2 sin theta cos theta)) = (2)/(g)`
(b) `4sqrt(H_(1)H_(2)) = 4sqrt((u^(2) sin^(2) theta)/(2g) xx (u^(2) cos^(2) theta)/(2g))`
`= 2(u^(2))/(g) sin theta cos theta`
`= (u^(2))/(g)(2 sin theta cos theta) = R`
`(4sqrt(H_(1)H_(2)))/(R ) = 1`