Show that the horizontal range of a projectile is same for two angles of projection . Is the sum of maximum heights for these two angles dependent on the angle of projectile ?

Explanation

When a projectile is fired at an angle `theta` with the horizontal , with a velocity u, its horizontal range is
`R=(u^(2) sin 2theta)/(g)`
Replacing `theta` by `(90^(@)-theta)`, we get
`R. =(u^(2) sin 2 (90^(@)-theta))/(g)=(u^(2) sin (180^(@)-2 theta)/(g))`
`=(u^(2) sin 2 theta)/(g)=R`
So for angles of projection `theta` and `90 ^(@)-theta`, horizontal range is same .
Maximum height attained by the projectile in two case is
`H=(u^(2) sin ^(2) theta)/(2g)`
and `H.=(u^(2) sin (90^(@)-theta))/(2g)=(u^(2) cos ^(2) theta)/(2g)`
Adding (i) and (ii) ,
`H+H.=(u^(2))/(2g)(sin^(2) theta+cos^(2) theta)=(u^(2))/(2g)`
The sum of maximum heights of projectile for these different velocities and different angle of projection .