a
(a)Shell is fired with velocity \(v\) at an angle \(\theta\) with the horizontal.
So its velocity at the highest point
\(=\) horizontal component of velocity \(=\)\(v\cos \theta \)
So momentum of shell before explosion \(=\) \(mv\cos \theta \)

When it breaks into two equal pieces and one piece retrace its path to the canon, then other part move with velocity \(V.\)

So momentum of two pieces after explosion
\( = \frac{m}{2}( - v\cos \theta ) + \frac{m}{2}V\)
By the law of conservation of momentum
\(mv\cos \theta = \frac{{ - m}}{2}v\cos \theta + \frac{m}{2}V\)==> \(V = 3v\cos \theta \)