Please use the second given kinematic equation to solve this

An archer releases an arrow from 1.3 rn above the ground. The arrow is released horizontally with velocity on. The center of the target is h = 0.70 rn above the ground and D = 55 rn away from the archer. What is the initial velocity, '00, in the horizontal direction needed for the arrow to hit the center of the target? Plan: mark the INCORRECT plan. Use 123 = '03\" 29(y yo) to find vy when the arrow hits the target; then use v;Ir = 1:0 gt to find t; (a) . . then use a:(t) = 2:0 + 0.3} to find '09,: then use no: to fund 120. (b) Use y(t) = yo + vent gt2 to calculate t;then use 305) = 2:0 + 'Uozt to find '00,? then use v.0: to find '00. (c) Use :1:(t) = 2:0 gt2 to calculate t; then use :r(t) = 9:0 + vozt to find '00,; then use Us: to find we. V100'l Result: mark the CORRECT result. (a) 0.13 m/s (b) 449.17 m/s (c) 157.13 m/s (d) 19.25 m/s (6} 222.23 mxs