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. (a)`40sqrt(3)` m
B. (b)40 m
C. (c)`20sqrt(3)` m
D. (d)20 m

Explanation

To solve the problem, we need to analyze the motion of the ball projected from the tower. We will break down the steps to find the x-coordinate when the velocity of the ball becomes perpendicular to the initial velocity of projection.

### Step 1: Determine the initial velocity components
The initial velocity \( v_0 \) of the ball is given as 20 m/s at an angle of \( 30^\circ \) with the horizontal.

- The horizontal component of the initial velocity \( v_{0x} \) is:
\[ v_{0x} = v_0 \cdot \cos(30^\circ) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

- The vertical component of the initial velocity \( v_{0y} \) is:
\[ v_{0y} = v_0 \cdot \sin(30^\circ) = 20 \cdot \frac{1}{2} = 10 \, \text{m/s} \]

### Step 2: Write the equations of motion
The horizontal motion is uniform since there is no acceleration in the x-direction. The vertical motion is influenced by gravity.

- The horizontal position \( x(t) \) as a function of time \( t \) is:
\[ x(t) = v_{0x} \cdot t = 10\sqrt{3} \cdot t \]

- The vertical position \( y(t) \) as a function of time \( t \) is:
\[ y(t) = v_{0y} \cdot t - \frac{1}{2} g t^2 = 10t - 5t^2 \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 3: Determine the velocity components at time \( t \)
The velocity components at any time \( t \) are:
- Horizontal velocity \( v_x(t) = v_{0x} = 10\sqrt{3} \, \text{m/s} \) (constant)
- Vertical velocity \( v_y(t) = v_{0y} - g t = 10 - 10t \)

### Step 4: Condition for perpendicular velocities
The velocities are perpendicular when the dot product of the initial velocity vector and the current velocity vector is zero:
\[ v_{0x} \cdot v_x(t) + v_{0y} \cdot v_y(t) = 0 \]
Substituting the values:
\[ (10\sqrt{3}) \cdot (10\sqrt{3}) + (10) \cdot (10 - 10t) = 0 \]
This simplifies to:
\[ 300 + 100 - 1000t = 0 \]
\[ 400 = 1000t \implies t = \frac{400}{1000} = 0.4 \, \text{s} \]

### Step 5: Calculate the x-coordinate at \( t = 0.4 \, \text{s} \)
Now, substitute \( t = 0.4 \) seconds into the equation for \( x(t) \):
\[ x(0.4) = 10\sqrt{3} \cdot 0.4 = 4\sqrt{3} \, \text{m} \]

### 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:
\[ \boxed{4\sqrt{3} \, \text{m}} \]