Kalpit and Mukesh are playing with two different balls of masses m and 2m respectively. If Kalpit throws his ball vertically up and Mukesh at an angle $\theta$,such that time of flight of both balls is same . The height attained by the two balls are in the ratio of : 

 Let ${\text{u}}_{\text{1}}$ and ${\text{u}}_{\text{2}}$ be the initial velocities respectively. If ${\text{h}}_{\text{1}}$ and ${\text{h}}_2$ are the

heights attained by them, then

${\text{h}}_{\text{1}} =\frac{u_1^2}{2g}{\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}h}}_{\text{2} }=\frac{u_2^2{\sin }^2\theta }{2g}\mathrm{...}\left(\text{1}\right)$

The times of ascent of balls are equal, we have t

t$={\text{u}}_{\text{1}}/\text{g\hspace{0.17em}}={\text{u}}_{\text{2}} \text{sin\hspace{0.17em}}\theta \text{/g}$

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

divide the heights of equation (1)

$\text{From eq}.\left(\text{1}\right)\frac{h_1}{h_2}=\frac{u_1^2}{u_2^2{\sin }^2\theta }\mathrm{...}\left(3\right)$

$$\begin{gathered} substitute equation\left(2\right) in \left(3\right) \\ \frac{h_1}{h_2}=\frac{1}{1} \end{gathered}$$