Given,
The range of the flight of the arrow, R=30.0 m
The initial launch velocity of the arrow, u=110.0 m
It is given that the arrow is launched horizontally, thus the vertical component of the launch velocity of the arrow is 0 m/s.
The range of the flight of the arrow is given by,
[tex]R=\frac{u}{t}[/tex]Where t is the time of flight of the arrow.
On substituting the known values,
[tex]\begin{gathered} 30.0=\frac{110.0}{t} \\ \Rightarrow t=\frac{110.0}{30.0} \\ =3.67\text{ s} \end{gathered}[/tex]From the equation of motion, the drop in the height of the arrow during its entire is given by,
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]Where g is the acceleration due to gravity and u_y is the vertical component of the initial velocity of the arrow.
On substituting the known values,
[tex]\begin{gathered} h=0+\frac{1}{2}\times9.8\times3.67^2 \\ =66\text{ m} \end{gathered}[/tex]Thus the arrow will hit 66 m below the bullseye.