Question

A boy throws a ball with a velocity \(V_0\) at an angle \(\alpha\) to the ground. At the same time, 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:

• A. \( V_0 \cos \alpha \)
• B. \( V_0 \sin \alpha \)
• C. \( V_0 \tan \alpha \)
• D. \( \sqrt{V_0^2 \tan \alpha} \)

Hint:

For a projectile, the horizontal component of velocity remains constant. If an object is thrown at an angle and another object moves uniformly to catch it, the required velocity of the moving object must match the horizontal component of the projectile’s velocity.

Answer 1

The Correct Option is A

Step 1: Understanding the motion of the projectile
The ball is projected with an initial velocity \( V_0 \) at an angle \( \alpha \) to the horizontal. The motion of the ball can be divided into two components: - Horizontal component: \( V_x = V_0 \cos \alpha \) - Vertical component: \( V_y = V_0 \sin \alpha \) The time taken for the ball to reach the ground is given by: \[ t = \frac{2 V_0 \sin \alpha}{g} \] Step 2: Distance traveled by the ball in horizontal direction
The horizontal range covered by the ball is: \[ R = V_x \cdot t = V_0 \cos \alpha \times \frac{2 V_0 \sin \alpha}{g} \] Step 3: Velocity required for the boy
Since the boy starts running at the same time and must reach the landing point of the ball, he must cover the same horizontal distance \( R \) in the same time \( t \). Thus, his velocity must be: \[ V_{\text{boy}} = \frac{R}{t} = V_0 \cos \alpha \] Step 4: Conclusion
Thus, the required velocity for the boy to catch the ball before it hits the ground is: \[ \boxed{V_0 \cos \alpha} \]