The velocity of the particle M at the instant of detachment, $\vec{v}_M$, is the vector sum of the velocity of the wheel's center, $\vec{v}_c$, and the velocity of the particle relative to the center, $\vec{v}_{rel}$.

1. Initial Velocity of Particle M The velocity of the center is purely horizontal:

   $$\vec{v}_c = v_c \hat{i}$$

   The particle is at point A, on the left side of the wheel at the same height as the center. The velocity relative to the center is tangential and points vertically upwards. Its magnitude is $v_{rel} = \omega R$.

   $$\vec{v}_{rel} = (\omega R) \hat{j}$$

   For a wheel rolling without slip, the condition $v_c = \omega R$ holds. Therefore, $v_{rel} = v_c$.

   $$\vec{v}_{rel} = v_c \hat{j}$$

   The total initial velocity of particle M is:

   $$\vec{v}_M = \vec{v}_c + \vec{v}_{rel} = v_c \hat{i} + v_c \hat{j}$$

   The initial velocity components are $v_{Mx} = v_c$ and $v_{My} = v_c$.

2. Projectile Motion and Time of Flight After detaching, particle M follows a projectile path. It starts at an initial height $h = R$ above the ground. We can find the time of flight, $t_{flight}$, by analyzing the vertical motion until it hits the ground ($y_{final} = 0$).

   $$y_{final} = y_{initial} + v_{My} t - \frac{1}{2}gt^2$$ $$0 = R + v_c t_{flight} - \frac{1}{2}g t_{flight}^2$$

   Rearranging gives a quadratic equation for $t_{flight}$:

   $$\frac{1}{2}g t_{flight}^2 - v_c t_{flight} - R = 0$$

   Solving for $t_{flight}$ and taking the physically meaningful positive root:

   $$t_{flight} = \frac{v_c + \sqrt{(-v_c)^2 - 4(\frac{g}{2})(-R)}}{2(\frac{g}{2})} = \frac{v_c + \sqrt{v_c^2 + 2gR}}{g}$$

3. Horizontal Distance The horizontal distance $D$ covered by the particle is the product of its constant horizontal velocity and the time of flight.

   $$D = v_{Mx} t_{flight} = v_c t_{flight}$$

   Substituting the expression for $t_{flight}$:

   $$D = v_c \left( \frac{v_c + \sqrt{v_c^2 + 2gR}}{g} \right) = \frac{v_c^2 + v_c\sqrt{v_c^2 + 2gR}}{g}$$

4. Calculation Substituting the given values $R = 0.55\text{ m}$, $v_c = 5\text{ m/s}$, and using $g = 9.8\text{ m/s}^2$:

   $$D = \frac{(5)^2 + 5\sqrt{(5)^2 + 2(9.8)(0.55)}}{9.8}$$ $$D = \frac{25 + 5\sqrt{25 + 10.78}}{9.8} = \frac{25 + 5\sqrt{35.78}}{9.8}$$ $$D \approx \frac{25 + 5(5.982)}{9.8} = \frac{25 + 29.91}{9.8} = \frac{54.91}{9.8}$$ $$D \approx 5.603\text{ m}$$