For given value of u, there are two angles of projection for which the horizontal range is the same. Show that the sum of the maximum heights for these two angles is independent of the angle of projection.

Explanation

If a projectile id projected with velocity (u0, making an theta with the morizontal direction, then
Horizontal range, ` R=(u^(2) /R sin 2 theta`
and Max. height , ` H=(u^(2) sin ^(2) thate )/ 2 g)`
Case (i) If theta =alph., let `R=R_(1) and H=H_(1)`
the `R_(1) =(u^(2) sin 2 alph)/g ...(i) and `H_(i) =u^(2) /2 g) sin^(2) alpha` ...(ii)
Case (ii) If ` theat =(90^(@)- alpha),` let ` R=R_(2) and H=H_(2), Then
` R_(1) =(u^(2) din 2 (90^(@)-alpha))/g =u^(2)/g sin (180^(@) -2 alpha)`
=` u^(2)/g sin 2 alpha` ...,..(iii)
`H_(2) =u^(2)/ 2g) sin^(2) (0^(@)-alpha) =u^(2)/(2 g) cos^(2) alpha` ...(iv) from (i) and (iii),
` R _(1) =R_(2)`
From 9ii) and 9iv),
` H_(1) + H_(2) =(u^(2)/(2 g) (sin ^(2) theta+ cos^(2)) =u^(2)/(2 g)`
Which is indepenednt of angle or projection ` theta` Thus proved.