Step 1: Understanding the Concept:
This problem can be solved using the principle of conservation of mechanical energy. The total mechanical energy (sum of kinetic and potential energy) of the girl on the swing is constant, assuming no air resistance or friction. The maximum velocity occurs at the lowest point of the swing, where the potential energy is minimum. The velocity is zero at the highest point, where the potential energy is maximum.
Step 2: Key Formula or Approach:
Conservation of Mechanical Energy:
\[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \]
where $KE = \frac{1}{2}mv^2$ and $PE = mgh$.
Let the initial point be the maximum height and the final point be the lowest point.
Step 3: Detailed Explanation:
Let's define the states:
- Initial state (highest point):
- Height, $h_{initial} = 3$ m.
- At the maximum height, the swing momentarily stops, so velocity $v_{initial} = 0$.
- Final state (lowest point):
- Height, $h_{final} = 2$ m.
- The velocity is maximum at this point, let it be $v_{max}$.
Apply the conservation of energy principle:
\[ \frac{1}{2}mv_{initial}^2 + mgh_{initial} = \frac{1}{2}mv_{max}^2 + mgh_{final} \]
Substitute the known values:
\[ \frac{1}{2}m(0)^2 + mg(3) = \frac{1}{2}mv_{max}^2 + mg(2) \]
\[ 3mg = \frac{1}{2}mv_{max}^2 + 2mg \]
The mass 'm' cancels out from all terms:
\[ 3g = \frac{1}{2}v_{max}^2 + 2g \]
Rearrange to solve for $v_{max}^2$:
\[ \frac{1}{2}v_{max}^2 = 3g - 2g = g \]
\[ v_{max}^2 = 2g \]
\[ v_{max} = \sqrt{2g} \]
Using the standard value of $g = 9.8 \text{ m/s}^2$:
\[ v_{max} = \sqrt{2 \times 9.8} = \sqrt{19.6} \text{ m/s} \]
Step 4: Final Answer:
The maximum velocity of the girl is $\sqrt{19.6}$ m/s. Therefore, option (C) is correct.