A ball is thrown from a point with a speed 'v^(0)' at an elevation angle of `theta` . From the same point and at the same instant , a person starts running with a constant speed `('v_(0)')/(2) ` to catch the ball . Will the person be able to catch the ball ? If yes, what should be the angle of projection `theta` ?

Options

A. No
B. yes, `30(@)`
C. `yes, 60(@)`
D. yes ` , 45(@)`

Explanation

To determine whether the person can catch the ball and at what angle of projection \( \theta \), we can analyze the motion of both the ball and the person.

### Step-by-Step Solution:

1. **Understand the Motion of the Ball**:
- The ball is thrown with an initial speed \( v_0 \) at an angle \( \theta \).
- The horizontal component of the ball's velocity is given by:
\[ v_{0x} = v_0 \cos(\theta) \]
- The vertical component of the ball's velocity is:
\[ v_{0y} = v_0 \sin(\theta) \]

2. **Time of Flight of the Ball**:
- The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2 v_{0y}}{g} = \frac{2 v_0 \sin(\theta)}{g} \]
- Here, \( g \) is the acceleration due to gravity.

3. **Horizontal Distance Covered by the Ball**:
- The horizontal distance \( R \) covered by the ball during its flight is:
\[ R = v_{0x} \cdot T = (v_0 \cos(\theta)) \cdot \left(\frac{2 v_0 \sin(\theta)}{g}\right) \]
- Simplifying this gives:
\[ R = \frac{2 v_0^2 \sin(\theta) \cos(\theta)}{g} = \frac{v_0^2 \sin(2\theta)}{g} \]

4. **Motion of the Person**:
- The person runs with a constant speed of \( \frac{v_0}{2} \).
- The horizontal distance covered by the person in the same time \( T \) is:
\[ D = \left(\frac{v_0}{2}\right) T = \left(\frac{v_0}{2}\right) \left(\frac{2 v_0 \sin(\theta)}{g}\right) = \frac{v_0^2 \sin(\theta)}{g} \]

5. **Setting Distances Equal**:
- For the person to catch the ball, the distances must be equal:
\[ R = D \]
- Substituting the expressions we derived:
\[ \frac{v_0^2 \sin(2\theta)}{g} = \frac{v_0^2 \sin(\theta)}{g} \]
- Canceling \( \frac{v_0^2}{g} \) from both sides (assuming \( v_0 \neq 0 \)):
\[ \sin(2\theta) = \sin(\theta) \]

6. **Using the Identity for Sine**:
- We can use the identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ 2 \sin(\theta) \cos(\theta) = \sin(\theta) \]
- Rearranging gives:
\[ 2 \sin(\theta) \cos(\theta) - \sin(\theta) = 0 \]
- Factoring out \( \sin(\theta) \):
\[ \sin(\theta)(2 \cos(\theta) - 1) = 0 \]

7. **Finding Possible Angles**:
- This gives us two cases:
1. \( \sin(\theta) = 0 \) → \( \theta = 0^\circ \) (not a valid angle for projection)
2. \( 2 \cos(\theta) - 1 = 0 \) → \( \cos(\theta) = \frac{1}{2} \) → \( \theta = 60^\circ \)

### Conclusion:
The person will be able to catch the ball if the angle of projection \( \theta \) is \( 60^\circ \).