A cannon fires a shell with a velocity V at an angle θ relative to...

At the highest point of the trajectory, the vertical component of the velocity of the shell is zero. The velocity of the shell is purely horizontal and is given by $v = V \cos\theta$.

The momentum of the shell just before the explosion is $M V \cos\theta$, where $M$ is the mass of the shell.

After the explosion, the shell breaks into two equal pieces, each of mass $\dfrac{M}{2}$.

Since one of the pieces retraces its path back to the cannon, its velocity must be exactly reversed. Therefore, the velocity of this piece just after the explosion is $-V \cos\theta$.

Let the velocity of the second piece be $v_2$. By the principle of conservation of linear momentum in the horizontal direction:

$M V \cos\theta = \dfrac{M}{2}(-V \cos\theta) + \dfrac{M}{2}v_2$

Dividing the entire equation by $M$:

$V \cos\theta = -\dfrac{V \cos\theta}{2} + \dfrac{v_2}{2}$

Multiplying by $2$:

$2V \cos\theta = -V \cos\theta + v_2$

$v_2 = 3V \cos\theta$

Since the vertical momentum is also conserved and the first piece has zero vertical velocity, the second piece must also have zero vertical velocity. Thus, the speed of the other piece immediately following the explosion is $3V \cos\theta$.