Answer
(a) The center of mass will be a distance of 32.3 meters from the launch point.
(b) The other piece lands a distance of 38.6 meters from the launch point.
Work Step by Step
(a) $range = \frac{v_0^2~sin(2\theta)}{g}$
$range = \frac{(18.0~m/s)^2~sin[(2)(51.0^{\circ})]}{9.80~m/s^2}$
$range = 32.3~m$
The center of mass will be a distance of 32.3 meters from the launch point.
(b) Since both pieces have an equal mass, they will both land the same distance from the center of mass. If one piece lands at a distance of 26.0 meters from the launch point, then the other piece also lands a distance of 6.3 meters from the center of mass, which is a distance of 38.6 meters from the launch point.
