A girl in a swing is 2.5m above ground at the maximum height and at 1.5m above the ground at the lowest point. Her maximum velocity in the swing is `(g = 10ms^(-2))`

Options

A. `5sqrt2 ms^(-1)`
B. `2sqrt5 ms^(-1)`
C. `2sqrt3 ms^(-1)`
D. `3sqrt2 ms^(-1)`

Explanation

To solve the problem of finding the maximum velocity of a girl on a swing, we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Identify the heights
- The maximum height of the swing above the ground is given as \( h = 2.5 \, \text{m} \).
- The lowest point of the swing above the ground is given as \( h' = 1.5 \, \text{m} \).

### Step 2: Determine the change in height
- The change in height (\( \Delta h \)) when the swing moves from the lowest point to the maximum height can be calculated as:
\[ \Delta h = h - h' = 2.5 \, \text{m} - 1.5 \, \text{m} = 1.0 \, \text{m} \]

### Step 3: Apply the conservation of energy
- At the lowest point, all the energy is kinetic, and at the maximum height, all the energy is potential. Therefore, we can set up the equation:
\[ \text{K.E. at lowest point} = \text{P.E. at maximum height} \]
- The kinetic energy (K.E.) at the lowest point is given by:
\[ \text{K.E.} = \frac{1}{2} m v_{\text{max}}^2 \]
- The potential energy (P.E.) at the maximum height is given by:
\[ \text{P.E.} = mgh \]
- Thus, we can write:
\[ \frac{1}{2} m v_{\text{max}}^2 = mg \Delta h \]

### Step 4: Simplify the equation
- We can cancel the mass \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v_{\text{max}}^2 = g \Delta h \]
- Rearranging gives:
\[ v_{\text{max}}^2 = 2g \Delta h \]

### Step 5: Substitute the known values
- We know \( g = 10 \, \text{m/s}^2 \) and \( \Delta h = 1.0 \, \text{m} \):
\[ v_{\text{max}}^2 = 2 \times 10 \, \text{m/s}^2 \times 1.0 \, \text{m} = 20 \, \text{m}^2/\text{s}^2 \]

### Step 6: Calculate \( v_{\text{max}} \)
- Taking the square root of both sides gives:
\[ v_{\text{max}} = \sqrt{20} = 2\sqrt{5} \, \text{m/s} \]

### Final Answer
- The maximum velocity of the girl in the swing at the lowest point is:
\[ v_{\text{max}} = 2\sqrt{5} \, \text{m/s} \]

---