Prove that for a given velocity of projection, the horizontal range is same for two angles of projection alpha and (90^(@) - alpha).

Prove that for a given velocity of projection, the horizontal range is

1.	Prove that for a given velocity of projection, the horizontal range is same for two angles of projection alpha and (90^(@) - alpha).
Answer» Solution :The horizontal rangeis given by `R = (u^(2) sin 2 THETA)/( g) "". . . (1)` when `theta = alpha ` `R_(1)= (u^(2) sin 2 alpha)/( g) "". . . (2)` when`theta = (90^(@) - alpha)` `R_(2) = (u^(2) sin 2 (90^(@) - alpha))/( g)` ` = (u^(2) 2 sin (90^(@) - alpha) cos (90^(@) - alpha))/(g) "". . . (3)` But `sin (90^(@) - alpha) = cos alpha , cos (90^(@) - alpha) = sin alpha ` `R_(2) = (u^(2) (2 sin alpha cos alpha))/( g)` ` = (u^(2) sin 2 alpha)/( g) "" . . . (4)` From (2) and (4), it is SEEN that at both angles `alpha and (90 - alpha)`, the horizontal RANGE remains the same .