Answer
The initial speed is
$$v_0=46\text{ m/s}.$$
Work Step by Step
We immediately see that the arrow has traveled $d_v=52\text{ cm}=0.52\text{ m}$ vertically downwards with no initial velocity in that direction which means that
$$d_v=\frac{1}{2}gt^2\Rightarrow t=\sqrt{\frac{2 d_v}{g}},$$
whete $t$ is the time of flight.
Now, we know that the arrow covers $d_h=15\text{ m}$ in the horizontal direction moving with the constant speed $v_0$, which is also the magnitude of the initial velocity, so we have:
$$d_h=v_0 t=v_0\sqrt{\frac{2d_v}{g}}$$
which yields
$$v_0=\sqrt{\frac{g}{2d_v}}d_h=\sqrt{\frac{9.81\text{ m/s}^2}{2\cdot0.52\text{ m}}}\cdot 15\text{ m}=46\text{ m/s}.$$
