Here, maximum height, $H=3 \mathrm{~m}$ and range, $R=4 \mathrm{~m}$

As, maximum height, $H=\frac{u^2 \sin ^2 \theta}{2 g}$ and maximum range, $R=\frac{u^2 \sin 2 \theta}{g}$



The ratio,
$$
\begin{array}{rlrl}
& & \frac{H}{R} & =\frac{3}{4}=\frac{\sin ^2 \theta}{2 \sin 2 \theta} \Rightarrow 3 \sin 2 \theta=2 \sin ^2 \theta \\
\Rightarrow \quad & \tan \theta & =3 \quad(\because \sin 2 \theta=2 \sin \theta \cos \theta)
\end{array}
$$

From Eqs. (i) and (ii), we get
$$
\begin{aligned}
& 3 \times 2 \times 10=u^2 \sin ^2 \theta \Rightarrow 3 \times 2 \times 10=u^2 \times\left(\frac{3}{\sqrt{10}}\right)^2 \\
& u^2=\left(\frac{\sqrt{10}}{3}\right)^2 \times 3 \times 2 \times 10 \Rightarrow u^2=\frac{100 \times 2}{3} \Rightarrow u=10 \sqrt{\frac{2}{3}}
\end{aligned}
$$

Hence, the correct option is (c).