(b)

As ranges are equal, so the angles of projection of two balls must be complementary.

$\text { i.e., } \theta_1+\theta_2 =90^{\circ}$

$\text { Given, } \theta_1 =30^{\circ}$

$\therefore \theta_2=90^{\circ}-30^{\circ}=60^{\circ}$

Maximum height of projectile (ball),

$H_{\max }=\frac{u^2 \sin ^2 \theta}{2 g}$

$\therefore \quad \frac{H_1}{H_2}=\frac{u_1^2 \sin ^2 \theta_1}{2 g} \times \frac{2 g}{u_1^2 \sin ^2 \theta_2}$

$=\frac {\sin ^2\left(30^{\circ}\right)}{\sin ^2\left(60^{\circ}\right)}$ $\left(\because u_1=u_2\right)$

$\frac{h}{H_2}=\frac{\left(\frac{1}{2}\right)^2}{\left(\frac{\sqrt{3}}{2}\right)^2}=\frac{1}{3} \quad\left(\because H_1=h\right)$

or $\quad H_2=3 h$