Given, speed of arrow,
$v=40 \mathrm{~m} / \mathrm{s}$ and $\theta=30^{\circ}$


Horizontal range $R_1$ covered by the arrow is given by
$\begin{aligned} R_1 & =\frac{v^2 \sin 2 \theta}{g} \\ & =\frac{40^2 \sin (2 \times 30)}{10}=160 \sin 60^{\circ} \\ & =160 \frac{\sqrt{3}}{2}=80 \sqrt{3} \mathrm{~m}\end{aligned}$
If $t$ be the time taken by the arrow reaching from $B$ to $C$, then from the second equation of motion,
$\Rightarrow \quad h=u t+\frac{1}{2} g t^2$
here, $h=4.2 \mathrm{~m}, u=v \sin 30^{\circ}$
$4.2=v \sin 30^{\circ} t+\frac{1}{2} g t^2$
$4.2=40 \times \frac{1}{2} t+\frac{1}{2} \times 10 t^2$
$4.2=20 t+5 t^2$
$\begin{aligned} & 50 t^2+200 t-42=0 \\ & 25 t^2+100 t-21=0\end{aligned}$
Solving the quadratic equation, $t=\frac{1}{5} \mathrm{~s}$ Distance travelled in horizontal direction is given by
$\begin{aligned} R_2 & =v \cos 30^{\circ} \times t \\ & =40 \frac{\sqrt{3}}{2} \times \frac{1}{5}=4 \sqrt{3} \mathrm{~m}\end{aligned}$
Hence, the horizontal distance covered by the arrow,
$\begin{aligned} R & =R_1+R_2 \\ & =80 \sqrt{3}+4 \sqrt{3}=84 \sqrt{3} \mathrm{~m}\end{aligned}$