An iron ball hangs from a 21.5m steel cable and is used in the demolition of a building at a location where the acceleration due to gravity is `9.75m//s^(2)`. The ball is swung outward from its equilibrium position for a distance of 4.20m. Assuming the system behaves as a simple pendulum, find the maximum speed of the ball during its swing.

Options

A. 4.45m/s
B. 2.83m/s
C. 17.8m/s
D. 1.71 m/s

Answer

To find the maximum speed of the iron ball during its swing, we can use the principle of conservation of mechanical energy. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the parameters**:
- Length of the steel cable (L) = 21.5 m
- Horizontal displacement (d) = 4.2 m
- Acceleration due to gravity (g) = 9.75 m/s²

2. **Determine the height (h) raised**:
- The height (h) to which the ball is raised can be calculated using the Pythagorean theorem. The vertical distance (x) from the pivot to the ball when swung out can be found as follows:
\[ x = \sqrt{L^2 - d^2} \]
Substituting the values:
\[ x = \sqrt{(21.5)^2 - (4.2)^2} = \sqrt{462.25 - 17.64} = \sqrt{444.61} \approx 21.1 \text{ m} \]
- Now, calculate the height (h) raised:
\[ h = L - x = 21.5 - 21.1 = 0.4 \text{ m} \]

3. **Apply the conservation of energy**:
- At the highest point of the swing, all the energy is potential energy (PE), and at the lowest point (equilibrium position), all the energy is kinetic energy (KE).
- Potential energy at the height (h):
\[ PE = mgh \]
- Kinetic energy at the lowest point:
\[ KE = \frac{1}{2} mv^2 \]
- By conservation of energy, we set PE equal to KE:
\[ mgh = \frac{1}{2} mv^2 \]
- The mass (m) cancels out:
\[ gh = \frac{1}{2} v^2 \]

4. **Solve for maximum speed (v)**:
- Rearranging the equation gives:
\[ v^2 = 2gh \]
- Taking the square root:
\[ v = \sqrt{2gh} \]
- Substituting the values:
\[ v = \sqrt{2 \times 9.75 \times 0.4} = \sqrt{7.8} \approx 2.8 \text{ m/s} \]

5. **Conclusion**:
- The maximum speed of the ball during its swing is approximately **2.83 m/s**.