A ball is projected from a high tower with speed 20 m/s at an angle `30^@` with horizontal x-axis. The x-coordinate of the ball at the instant when the velocity of the ball becomes perpendicular to the velocity of projection will be (take point of projection as origin):

Options

A. `40sqrt(3)` m
B. 40 m
C. `20sqrt(3)` m
D. 20 m

Explanation

To solve the problem step by step, we will analyze the motion of the ball projected from a tower. The key points to consider are the initial velocity components, the equations of motion, and the condition for the velocity vector to be perpendicular to the initial velocity vector.

### Step 1: Determine the initial velocity components
The ball is projected with a speed of 20 m/s at an angle of 30 degrees with the horizontal. We can break this initial velocity into its horizontal (Ux) and vertical (Uy) components using trigonometric functions.

- \( U_x = U \cdot \cos(\theta) = 20 \cdot \cos(30^\circ) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \)
- \( U_y = U \cdot \sin(\theta) = 20 \cdot \sin(30^\circ) = 20 \cdot \frac{1}{2} = 10 \, \text{m/s} \)

### Step 2: Write the equations of motion
Since there is no horizontal acceleration (Ax = 0) and the vertical acceleration is due to gravity (Ay = -g), we can write the equations of motion for the x and y coordinates.

- Horizontal motion: \( x(t) = U_x \cdot t = 10\sqrt{3} \cdot t \)
- Vertical motion: \( y(t) = U_y \cdot t - \frac{1}{2} g t^2 = 10t - \frac{1}{2} g t^2 \)

### Step 3: Determine the velocity components
The velocity components at any time t can be expressed as:

- \( V_x = U_x = 10\sqrt{3} \, \text{m/s} \) (constant)
- \( V_y = U_y - g \cdot t = 10 - g \cdot t \)

### Step 4: Condition for perpendicularity
The velocity vector becomes perpendicular to the initial velocity vector when their dot product is zero:

\[ \vec{V} \cdot \vec{U} = 0 \]

Substituting the components:

\[ (10\sqrt{3}) \cdot (10\sqrt{3}) + (10 - g \cdot t) \cdot 10 = 0 \]

This simplifies to:

\[ 300 + 100 - 10g \cdot t = 0 \]

\[ 400 = 10g \cdot t \]

\[ t = \frac{400}{10g} = \frac{40}{g} \]

### Step 5: Calculate the time when the velocity is perpendicular
Using \( g \approx 10 \, \text{m/s}^2 \):

\[ t = \frac{40}{10} = 4 \, \text{s} \]

### Step 6: Calculate the x-coordinate at this time
Now we can find the x-coordinate when \( t = 4 \, \text{s} \):

\[ x(4) = U_x \cdot t = 10\sqrt{3} \cdot 4 = 40\sqrt{3} \, \text{m} \]

### Final Answer
The x-coordinate of the ball at the instant when the velocity of the ball becomes perpendicular to the velocity of projection is \( 40\sqrt{3} \, \text{m} \).

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