Answer:
Approximately [tex]0.077\; {\rm m\cdot s^{-1}}[/tex] (assuming that external forces on the cannon are negligible.)
Explanation:
If an object of mass [tex]m[/tex] is moving at a velocity of [tex]v[/tex], the momentum [tex]p[/tex] of that object would be [tex]p = m\, v[/tex].
Momentum of the t-shirt:
[tex]\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}[/tex].
If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if [tex]p(\text{cannon})[/tex] denote the momentum of this cannon:
[tex]p(\text{t-shirt}) + p(\text{cannon}) = 0[/tex].
[tex]p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
Rewrite [tex]p = m\, v[/tex] to obtain [tex]v = (p / m)[/tex]. Since the mass of this cannon is [tex]m(\text{cannon}) = 33\; {\rm kg}[/tex], the velocity of this cannon would be:
[tex]\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].
