#### Answer

The initial speed of the arrow is 106 m/s

#### Work Step by Step

When the arrow hits the ground: $t = \frac{v_y}{g}$
When the arrow hits the ground:
$\frac{v_y}{v_x} = tan(3.0^{\circ})$
$v_y = v_x~tan(3.0^{\circ})$
We can find the initial speed of the arrow $v_x$.
$v_x~t = x$
$\frac{v_x~v_y}{g} = x$
$\frac{v_x^2~tan(3.0^{\circ})}{g} = x$
$v_x^2 = \frac{gx}{tan(3.0^{\circ})}$
$v_x = \sqrt{\frac{gx}{tan(3.0^{\circ})}}$
$v_x = \sqrt{\frac{(9.80~m/s^2)(60~m)}{tan(3.0^{\circ})}}$
$v_x = 106~m/s$
The initial speed of the arrow is 106 m/s.