A particle is projected with velocity u so that its horizontal...

$$\begin{aligned} & \text{ Range }=3 \mathrm{H}_{\max } \\ & \frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{ g}}=\frac{3 \mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{ g}}\end{aligned}$$2 \sin \theta \cos \theta=\frac{3}{2} \sin ^2 \theta \tan \theta=\frac{4}{3} \Rightarrow \theta=53^{\circ} \mathrm{R}=\frac{\mathrm{u}^2\left(2 \times \frac{3}{5} \times \frac{4}{5}\right)}{\mathrm{g}} \Rightarrow \frac{24 \mathrm{u}^2}{25 \mathrm{ g}}$