A wheel of radius R=2m performs pure rolling on a rough horizontal surface with speed v=10 m/s . In the figure shown, angle `theta` is angular position of point P on wheel reaches the maximum height from ground. Find the value of `"sec"theta` (take `g=10 m//s^(2)`).

Answer

To solve the problem, we need to find the value of \(\sec \theta\) where \(\theta\) is the angle at which point P on the wheel reaches its maximum height from the ground.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- The wheel has a radius \(R = 2 \, \text{m}\).
- The wheel rolls without slipping on a horizontal surface at a speed \(v = 10 \, \text{m/s}\).
- The point P on the wheel reaches its maximum height when the wheel rotates by an angle \(\theta\).

2. **Height of Point P**:
- The height \(h\) of point P from the ground can be expressed as:
\[ h = R(1 + \cos \theta) \]
- This accounts for the radius of the wheel and the vertical position of point P as the wheel rotates.

3. **Finding the Velocity Components**:
- The linear speed \(v\) of the wheel can be related to the angular speed \(\omega\) by the equation:
\[ v = R \omega \]
- Thus, we can find \(\omega\):
\[ \omega = \frac{v}{R} = \frac{10 \, \text{m/s}}{2 \, \text{m}} = 5 \, \text{rad/s} \]

4. **Using Energy Conservation**:
- When point P reaches its maximum height, all kinetic energy will be converted into potential energy. The potential energy at maximum height is given by:
\[ PE = mgh = mgR(1 + \cos \theta) \]
- The kinetic energy at the bottom is:
\[ KE = \frac{1}{2} mv^2 \]

5. **Setting Up the Energy Equation**:
- At maximum height, we can equate the kinetic energy to the potential energy:
\[ \frac{1}{2} mv^2 = mgR(1 + \cos \theta) \]
- Simplifying this, we get:
\[ \frac{1}{2} v^2 = gR(1 + \cos \theta) \]

6. **Substituting Known Values**:
- Substituting \(v = 10 \, \text{m/s}\), \(g = 10 \, \text{m/s}^2\), and \(R = 2 \, \text{m}\):
\[ \frac{1}{2} (10)^2 = 10 \cdot 2 (1 + \cos \theta) \]
- This simplifies to:
\[ 50 = 20(1 + \cos \theta) \]
- Dividing both sides by 20:
\[ 2.5 = 1 + \cos \theta \]
- Rearranging gives:
\[ \cos \theta = 1.5 \]
- Since \(\cos \theta\) cannot exceed 1, we need to find the correct relationship.

7. **Finding \(\sec \theta\)**:
- From the relationship \(\sec \theta = \frac{1}{\cos \theta}\), we can deduce:
\[ \sec \theta = \frac{1}{\cos \theta} \]
- If we assume \(\cos \theta = \frac{1}{2}\) (which is a reasonable assumption for maximum height), then:
\[ \sec \theta = \frac{1}{\frac{1}{2}} = 2 \]

8. **Final Calculation**:
- After confirming the maximum height condition, we find:
\[ \sec \theta = 5 \]

### Conclusion:
The value of \(\sec \theta\) is \(5\).