Pankaj and Sudhir are playing with two different balls of masses m and 2m, respectively. If Pankaj throws his ball vertically up and Sudhir at an angle $\theta with the verticle$, both of them stay
in our view for the same period. The height attained by the two balls are in the ratio

Time of flight for the ball thrown by Pankaj, $T_1=\frac{2u_1}{g}$

Time of flight for the ball thrown by Sudhir

$T_2=\frac{2u_2 \sin \left({90}^{\circ }-\theta \right)}{g}=\frac{2{u_2}^{ }\cos \theta }{g}$

According to problem $T_1=T_2$

$\begin{array}{l}\Rightarrow \frac{2u_1}{g}=\frac{2u_2\cos \theta }{g}\\ \Rightarrow u_1=u_2\cos \theta \end{array}$

Height of the ball thrown by pankaj $H_1=\frac{u_1^2}{2g}$
Height of the ball thrown by Sudhir

$\begin{array}{l}{H}_2=\frac{u_2^2{\sin }^2\left({90}^{\circ }-\theta \right)}{2g}=\frac{u_2^2{\cos }^2\theta }{2g}\\ \therefore \frac{{\mathrm{H}}_1}{H_2}=\frac{u_1^2/2g}{u_2^2{\cos }^2\theta /2g}=1 \left[\text{ As }{u}_1=u_2\cos \theta \right]\end{array}$