A child is swinging a swing. Minimum and maximum heights fo swing from the earth's surface are 0.75 m and 2 m respectively. The maximum velocity of this swing is

Options

A. 10 m/s
B. 5 m/s
C. 8 m/s
D. 15 m/s

Explanation

To find the maximum velocity of the swing, we can use the principle of conservation of mechanical energy. The potential energy at the maximum height will be converted into kinetic energy at the lowest point of the swing.

### Step-by-Step Solution:

1. **Identify the heights:**
- Minimum height (h1) = 0.75 m
- Maximum height (h2) = 2 m

2. **Calculate the change in height (Δh):**
\[ \Delta h = h2 - h1 = 2 \, \text{m} - 0.75 \, \text{m} = 1.25 \, \text{m} \]

3. **Apply the conservation of energy:**
The potential energy at the maximum height (h2) will be equal to the kinetic energy at the minimum height (h1).
\[ mgh2 = \frac{1}{2} mv_{\text{max}}^2 \]
Here, \( m \) is the mass of the child (which will cancel out), \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( v_{\text{max}} \) is the maximum velocity.

4. **Rearranging the equation:**
Since we are interested in \( v_{\text{max}} \), we can rearrange the equation:
\[ v_{\text{max}}^2 = 2g(h2 - h1) \]

5. **Substituting the values:**
\[ v_{\text{max}}^2 = 2g(1.25) \]
\[ v_{\text{max}}^2 = 2 \times 9.81 \times 1.25 \]

6. **Calculating \( v_{\text{max}}^2 \):**
\[ v_{\text{max}}^2 = 24.525 \, \text{m}^2/\text{s}^2 \]

7. **Finding \( v_{\text{max}} \):**
\[ v_{\text{max}} = \sqrt{24.525} \approx 4.95 \, \text{m/s} \]

8. **Final result:**
The maximum velocity of the swing is approximately \( 5 \, \text{m/s} \).