A projectile of mass M is fired from the ground with an initial...

At the highest point of the trajectory, the vertical component of velocity is zero. The horizontal component of velocity is:
$v_x = u \cos 60^{\circ} = \dfrac{u}{2}$

The kinetic energy of the projectile just before the explosion is:
$K_i = \dfrac{1}{2} M \left(\dfrac{u}{2}\right)^2 = \dfrac{M u^2}{8}$

The momentum of the projectile just before the explosion is entirely horizontal:
$p_i = M \left(\dfrac{u}{2}\right) = \dfrac{M u}{2}$

Let $v_f$ be the velocity of the heavier fragment (mass $\dfrac{2M}{3}$) after the explosion. Since the lighter fragment (mass $\dfrac{M}{3}$) comes to rest, its momentum is zero.

By conservation of linear momentum along the horizontal direction:
$p_i = p_f$
$\dfrac{M u}{2} = \left(\dfrac{M}{3}\right)(0) + \left(\dfrac{2M}{3}\right) v_f$
$\implies v_f = \dfrac{3 u}{4}$

The kinetic energy of the system just after the explosion is:
$K_f = \dfrac{1}{2} \left(\dfrac{2M}{3}\right) \left(\dfrac{3 u}{4}\right)^2 = \left(\dfrac{M}{3}\right) \left(\dfrac{9 u^2}{16}\right) = \dfrac{3 M u^2}{16}$

The extra energy released during the explosion is the increase in kinetic energy:
$\Delta K = K_f - K_i = \dfrac{3 M u^2}{16} - \dfrac{M u^2}{8} = \dfrac{3 M u^2 - 2 M u^2}{16} = \dfrac{M u^2}{16}$

Answer: $\dfrac{M u^2}{16}$