An archer shoots an arrow horizontally at a target 12 m away. The arrow is aimed directly at the center of the target, but it hits 51 cm lower. What was the initial speed of the arrow?

Your mistake was to use v0 in the calculation of height. The problem statement says the arrow was shot horizontally, so the initial speed in the vertical direction is zero, while the initial speed in the horizontal direction is v0. The vertical drop is

d = (1/2)gt2 = 0.56 m

Get t from this, and then set

v0t = 15 m [i.e. the horizontal travel distance at constant speed],

and then get v0 knowing t.

(1/2)(9.8)t2 = 0.56

t = √[2(0.56)/9.8] s ≅ 0.338 s

v0t = 15 m

v0 = (15 m)/(0.338 s) ≅ 44.4 m/s in the horizontal direction, with no initial vertical velocity

mass and volume are the answers you are looking for

because if the tool that you are using is not balanced it there would be false mass.