Answer
The front of the boat should be a distance of 25.5 meters from the dock when the equipment is thrown.
Work Step by Step
We can find $v_{0y}$ when the equipment leaves the tower.
$v_{0y} = v_0~sin(\theta) = (15.0~m/s)~sin(60.0^{\circ})$
$v_{0y} = 12.99~m/s$
We can find $v_y$ when the equipment lands on the deck.
$v_y^2 = v_{0y}^2+2gy$
$v_y = \sqrt{v_{0y}^2+2gy}$
$v_y = \sqrt{(12.99~m/s)^2+(2)(-9.80~m/s^2)(-8.75~m)}$
$v_y = 18.45~m/s$ (toward the ground)
We can find the time $t$ for the equipment's flight in the air.
$t = \frac{v_y-v_{0y}}{g} = \frac{(-18.45~m/s)-(12.99~m/s)}{-9.80~m/s^2}$
$t = 3.208~s$
We can find the horizontal distance $x$ that the equipment travels.
$x = v_x~t = (15.0~m/s)~cos(60.0^{\circ})(3.208~s)$
$x = 24.1~m$
We can find the horizontal distance $d$ that the boat travels.
$d = v_x~t = (0.450~m/s)(3.208~s)$
$d = 1.44~m$
The total distance $D$ from the dock is x+d.
$D = x+d = 24.1~m + 1.44~m = 25.5~m$
The front of the boat should be a distance of 25.5 meters from the dock when the equipment is thrown.
