Answer
The lighter fragment will land a distance of 8630 meters from the launch point.
The amount of energy released in the explosion was 532500 J.
Work Step by Step
Since the two fragments reach the ground at the same time, they must be moving horizontally just after the explosion.
Since the heavier fragment returns to the launch point, the velocity of the heavier fragment must be $-v_{0x}$ just after the explosion. We can use conservation of momentum to find the velocity of the lighter fragment.
$(9.0~kg)(-v_{0x})+(3.0~kg)(v) = (12.0~kg)(v_{0x})$
$v = \frac{(12.0~kg)(v_{0x})+(9.0~kg)(v_{0x})}{3.0~kg}$
$v = \frac{(21.0~kg)(150~m/s)~cos(55.0^{\circ})}{3.0~kg}$
$v = 602~m/s$
We can find the time for the shell to reach maximum height.
$t = \frac{v_{0y}}{g} = \frac{(150~m/s)~sin(55.0^{\circ})}{9.80~m/s^2}$
$t = 12.54~s$
Note that the time to go up is equal to the time it takes to fall back down again.
$range = (150~m/s)~cos(55.0^{\circ})(12.54~s)+(602~m/s)(12.54~s)$
$range = 8630~m$
The lighter fragment will land a distance of 8630 meters from the launch point.
We can find the kinetic energy just before the explosion.
$K_1 = \frac{1}{2}mv_{0x}^2$
$K_1 = \frac{1}{2}(12.0~kg)[(150~m/s)~cos(55.0^{\circ})]^2$
$K_1 = 44400~J$
We can find the kinetic energy just after the explosion.
$K_1 = \frac{1}{2}(m_1)(-v_{0x})^2+\frac{1}{2}(m_2)v^2$
$K_1 = \frac{1}{2}(9.0~kg)[(-150~m/s)~cos(55.0^{\circ})]^2+\frac{1}{2}(3.0~kg)(602~m/s)^2$
$K_1 = 576900~J$
$\Delta K = 576900~J - 44400~J$
$\Delta K = 532500~J$
The amount of energy released in the explosion was 532500 J.
