A sphere is suspended by a thread of length l. What minimum horizontal velocity has to be imparted to the ball for it to reach the height of the suspension?

Options

A. `sqrt("5 gl")`
B. 2 gl
C. `sqrt(gl)`
D. `sqrt(2gl)`

Explanation

To solve the problem of finding the minimum horizontal velocity that must be imparted to a sphere suspended by a thread of length \( l \) for it to reach the height of suspension, we will use the principle of conservation of mechanical energy.

### Step-by-Step Solution:

1. **Identify the Initial and Final States:**
- The initial state (point A) is when the sphere is at the lowest point of its swing, and we will consider this point as our reference level for potential energy, where \( U_A = 0 \).
- The final state (point B) is when the sphere reaches the height of suspension (point O), where its potential energy will be maximum and kinetic energy will be minimum.

2. **Apply Conservation of Mechanical Energy:**
- According to the conservation of mechanical energy:
\[ U_A + K_A = U_B + K_B \]
- At point A:
- Potential Energy \( U_A = 0 \)
- Kinetic Energy \( K_A = \frac{1}{2} m u^2 \) (where \( u \) is the initial horizontal velocity imparted)
- At point B:
- Potential Energy \( U_B = mg h \) (where \( h = l \), the height of suspension)
- Kinetic Energy \( K_B = 0 \) (since we want to find the minimum velocity at the highest point, where the speed is momentarily zero)

3. **Set Up the Equation:**
- Substitute the energies into the conservation equation:
\[ 0 + \frac{1}{2} m u^2 = mg l + 0 \]
- This simplifies to:
\[ \frac{1}{2} m u^2 = mg l \]

4. **Solve for \( u^2 \):**
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} u^2 = g l \]
- Multiply both sides by 2:
\[ u^2 = 2 g l \]

5. **Find \( u \):**
- Taking the square root of both sides gives:
\[ u = \sqrt{2 g l} \]

### Final Answer:
The minimum horizontal velocity that must be imparted to the sphere is:
\[ u = \sqrt{2 g l} \]