A ball of mass m is thrown vertically upward. Another ball of mass...

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, ${T}_{1} = {T}_{2}$ $⇒ \dfrac{2{u}_{1}}{g} = \dfrac{2{u}_{2}sin\theta}{g}$

$⇒ {u}_{1} = {u}_{2}sin\theta . . . \left(1\right)$

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

Ratio of the heights is $\dfrac{{H}_{1}}{{H}_{2}} = {u}_{1}^{2}{u}_{2}^{2}\sin^{2}\theta . . .$\left(2\right)$</p><p>From equation \left(1\right)$ and \left(2\right)$, we have </p><p>$⇒ \dfrac{{H}_{1}}{{H}_{2}} = 1$.</p><p>Thus,$x = 1.