A ball was thrown by a boy a at angle `60^(@)` with horizontal at height 1 m from ground. Boy B is running in the plane of motion of ball and catches the ball at height 1 m from ground. He finds the ball falling vertically. If the boy is running at a speed 20 km/hr. Then the velocity of projection of ball is-

Options

A. 20 km/hr
B. 30 km/hr
C. 40 km/hr
D. 50 km/hr

Answer

To solve the problem step by step, we need to analyze the motion of the ball and the conditions given in the question.

### Step 1: Understand the scenario
A ball is thrown at an angle of 60 degrees with the horizontal from a height of 1 meter. Boy B is running horizontally at a speed of 20 km/hr and catches the ball at the same height of 1 meter. The ball appears to be falling vertically to Boy B when he catches it.

### Step 2: Identify the components of motion
The velocity of the ball can be broken down into two components:
- Horizontal component (\(V_x\))
- Vertical component (\(V_y\))

Given that the ball is thrown at an angle of 60 degrees, we can express these components as:
- \(V_x = V \cos(60^\circ)\)
- \(V_y = V \sin(60^\circ)\)

### Step 3: Relate the horizontal motion to Boy B's speed
Since Boy B catches the ball while running horizontally, the horizontal component of the ball's velocity must equal Boy B's speed. Therefore:
\[ V_x = 20 \text{ km/hr} \]

### Step 4: Substitute the horizontal component
From Step 2, we have:
\[ V \cos(60^\circ) = 20 \text{ km/hr} \]
We know that:
\[ \cos(60^\circ) = \frac{1}{2} \]
Substituting this value in, we get:
\[ V \cdot \frac{1}{2} = 20 \text{ km/hr} \]

### Step 5: Solve for the velocity of projection \(V\)
To find \(V\), we can rearrange the equation:
\[ V = 20 \text{ km/hr} \cdot 2 \]
\[ V = 40 \text{ km/hr} \]

### Conclusion
The velocity of projection of the ball is \(40 \text{ km/hr}\).

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