Two particles A and B are projected from the same point with the same speed but at different angles $\alpha$ and $\beta$ with the horizontal, such that the maximum height of  A is ${(2/3)}^{rd}$ of horizontal range of B. Then which of the following relations are true

$\frac{v^2{\sin }^2\alpha }{2g}=\frac{2}{3}\frac{v^2\sin 2\beta }{g}$

$3{\sin }^2\alpha =4\sin 2\beta$

$\Rightarrow {\sin }^2\alpha =\frac{4}{3}\sin 2\beta$...(1)

$0\le {\sin }^2\alpha \le 1 \mathrm{and} 0\le \sin 2\beta \le 1 \Rightarrow \beta \le {\sin }^{-1}\left(\frac{3}{4}\right)$

$R_A=\frac{v^2\sin 2\alpha }{g}$ 

The range of A will be maximum $\mathrm{\alpha }=\frac{\mathrm{\pi }}{4}$

Using (1) $\beta =\frac{1}{2}{\sin }^{-1}\left(\frac{3}{8}\right)$

$R_A=\frac{v^2\sin 2\alpha }{g}$, $H_{max.B}=\frac{v^2{\sin }^2\beta }{2g}$

on equating we get $0<\alpha <\frac{\mathrm{\pi }}{12}$, and from (1) we get some values of $\alpha$ which are not in this range, So the statement of range of A= maximum height of B is not always true.