A ball of mass m is thrown vertically upward. Another ball of mass 2   m is thrown an angle θ with…

 When the first ball of mass $m$ is projected vertically upwards, the second ball of mass $2m$ makes an angle $\theta$ with the vertical.

The time of flight is the same for both the balls.

Then, $T_1=T_2$ $\Rightarrow \frac{2u_1}{g}=\frac{2u_2\sin \theta }{g}$

$\Rightarrow u_1=u_2\sin \theta ...\left(1\right)$

Maximum height of projectile is given by $H=\frac{u^2{\sin }^2\theta }{2g}$

Ratio of the heights is $\frac{H_1}{H_2}=\frac{u_1^2}{u_2^2{\sin }^2\theta } ...\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we have 

$\Rightarrow \frac{H_1}{H_2}=1$.

Thus, $x=1$.