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 solve the problem of whether the person can catch the ball and at what angle of projection, we can follow these steps:

### Step 1: Understand the motion of the ball
The ball is thrown with an initial speed \( v_0 \) at an angle \( \theta \). The horizontal and vertical components of the ball's velocity can be expressed as:
- Horizontal component: \( v_{0x} = v_0 \cos \theta \)
- Vertical component: \( v_{0y} = v_0 \sin \theta \)

### Step 2: Analyze the motion of the person
The person starts running horizontally with a constant speed of \( \frac{v_0}{2} \).

### Step 3: Determine the time of flight of the ball
The time of flight \( T \) of the ball can be calculated using the vertical motion. The ball will return to the ground when the vertical displacement is zero. The time of flight for projectile motion is given by:
\[ T = \frac{2 v_{0y}}{g} = \frac{2 v_0 \sin \theta}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 4: Calculate the horizontal distance traveled by the ball
The horizontal distance \( R \) traveled by the ball when it lands can be calculated as:
\[ R = v_{0x} \cdot T = (v_0 \cos \theta) \cdot \left(\frac{2 v_0 \sin \theta}{g}\right) = \frac{2 v_0^2 \sin \theta \cos \theta}{g} \]

### Step 5: Calculate the distance covered by the person
The distance \( D \) covered by the person while the ball is in the air is:
\[ D = \text{speed} \cdot \text{time} = \left(\frac{v_0}{2}\right) \cdot T = \left(\frac{v_0}{2}\right) \cdot \left(\frac{2 v_0 \sin \theta}{g}\right) = \frac{v_0^2 \sin \theta}{g} \]

### Step 6: Set the distances equal for catching the ball
For the person to catch the ball, the horizontal distance traveled by the ball must equal the distance covered by the person:
\[ \frac{2 v_0^2 \sin \theta \cos \theta}{g} = \frac{v_0^2 \sin \theta}{g} \]

### Step 7: Simplify the equation
Cancelling \( \frac{v_0^2 \sin \theta}{g} \) from both sides (assuming \( v_0 \neq 0 \) and \( \sin \theta \neq 0 \)):
\[ 2 \cos \theta = 1 \]
\[ \cos \theta = \frac{1}{2} \]

### Step 8: Solve for \( \theta \)
The angle \( \theta \) that satisfies \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = 60^\circ \]

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