Answer
The horizontal distance to the target is $~~6.85~m$
Work Step by Step
In part (a), we found that the direction of the initial velocity is $\theta = 55.6^{\circ}$ above the horizontal.
We can find the time to reach maximum height:
$t = \frac{v_{0y}}{a_y}$
$t = \frac{(12.0~m/s)~sin~55.6^{\circ}}{9.8~m/s^2}$
$t = 1.01~s$
We can find the horizontal distance to the target:
$x = v_x~t$
$x = (12.0~m/s)~(cos~55.6^{\circ})~(1.01~s)$
$x = 6.85~m$
The horizontal distance to the target is $~~6.85~m$
