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. Yes, `60^(@)`
B. Yes, `30^(@)`
C. No
D. Yes, `45^(@)`

Explanation

To determine whether the person can catch the ball and at what angle of projection \( \theta \), we can follow these steps:

### Step 1: Identify the horizontal 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:
\[ v_{x} = v_0 \cos \theta \]

### Step 2: Determine the time of flight
The time of flight \( t \) for the ball can be expressed in terms of its vertical motion. However, since we are interested in the horizontal distance, we can focus on the horizontal distance traveled during this time.

### Step 3: Calculate the horizontal distance traveled by the ball
The horizontal distance traveled by the ball in time \( t \) is:
\[ \text{Distance}_{\text{ball}} = v_{x} \cdot t = v_0 \cos \theta \cdot t \]

### Step 4: Identify the motion of the person
The person starts running with a constant speed of \( \frac{v_0}{2} \). The distance traveled by the person in the same time \( t \) is:
\[ \text{Distance}_{\text{person}} = \left(\frac{v_0}{2}\right) \cdot t \]

### Step 5: Set the distances equal for catching the ball
For the person to catch the ball, the distances traveled by both the ball and the person must be equal:
\[ v_0 \cos \theta \cdot t = \left(\frac{v_0}{2}\right) \cdot t \]

### Step 6: Simplify the equation
Since \( t \) is common on both sides, we can cancel it out (assuming \( t \neq 0 \)):
\[ v_0 \cos \theta = \frac{v_0}{2} \]

### Step 7: Solve for \( \cos \theta \)
Dividing both sides by \( v_0 \) (assuming \( v_0 \neq 0 \)):
\[ \cos \theta = \frac{1}{2} \]

### Step 8: Find the angle \( \theta \)
The angle \( \theta \) can be found using the inverse cosine function:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \]
This gives:
\[ \theta = 60^\circ \]

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

---