A boy throws a ball with a velocity u at an angle `theta` with the horizontal. At the same instant he starts running with uniform velocity to catch the ball before it hits the ground. To achieve this he should run with a velocity of

Options

A. `u cos theta`
B. `u sin theta`
C. `u tan theta`
D. `u sec theta`

Explanation

To solve the problem step by step, we need to analyze the motion of the ball and the boy running to catch it.

### Step 1: Understand the motion of the ball
When the boy throws the ball with an initial velocity \( u \) at an angle \( \theta \) with the horizontal, we can break this velocity into two components:
- **Horizontal component**: \( u_x = u \cos \theta \)
- **Vertical component**: \( u_y = u \sin \theta \)

### Step 2: Determine the time of flight of the ball
The time of flight \( T \) for a projectile launched at an angle \( \theta \) can be calculated using the vertical component of the velocity. The time of flight until the ball hits the ground is given by the formula:
\[ T = \frac{2u_y}{g} = \frac{2u \sin \theta}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 3: Calculate the horizontal distance traveled by the ball
During the time \( T \), the horizontal distance \( R \) traveled by the ball can be calculated using the horizontal component of the velocity:
\[ R = u_x \cdot T = (u \cos \theta) \cdot \left(\frac{2u \sin \theta}{g}\right) \]
This simplifies to:
\[ R = \frac{2u^2 \sin \theta \cos \theta}{g} \]

### Step 4: Determine the required velocity of the boy
To catch the ball, the boy must cover the same horizontal distance \( R \) in the same time \( T \). If the boy runs with a constant velocity \( v \), the distance he covers is:
\[ \text{Distance covered by boy} = v \cdot T \]
Setting this equal to the distance traveled by the ball gives:
\[ v \cdot T = R \]
Substituting \( T \) and \( R \) into this equation, we get:
\[ v \cdot \left(\frac{2u \sin \theta}{g}\right) = \frac{2u^2 \sin \theta \cos \theta}{g} \]

### Step 5: Solve for the velocity \( v \)
Cancelling \( \frac{2\sin \theta}{g} \) from both sides (assuming \( \sin \theta \neq 0 \)):
\[ v = u \cos \theta \]

### Final Answer
Thus, the velocity with which the boy should run to catch the ball is:
\[ \boxed{u \cos \theta} \]

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