A particle is projected from a point (0,1) of Y-axis (assume + Y direction vertically upwards) aiming towards a point (4,9). It fell on ground along x axis in 1 sec. Taking `g=10(m)/(s^2)` and all coordinate in metres. Find the X-coordinate where it fell.

Options

A. `93,0)`
B. `(4,0)`
C. `(2,0)`
D. `(2sqrt(5),0)`

Explanation

To solve the problem step by step, we will follow the physics of projectile motion and the given conditions.

### Step 1: Understand the Problem
A particle is projected from the point (0, 1) aiming towards the point (4, 9) and falls on the ground (x-axis) after 1 second. We need to find the x-coordinate where it fell.

### Step 2: Determine the Slope
To find the angle of projection, we first calculate the slope of the line connecting the points (0, 1) and (4, 9).

\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{9 - 1}{4 - 0} = \frac{8}{4} = 2 \]

Thus, we have:

\[ \tan \theta = 2 \]

### Step 3: Use the Time of Flight
The time of flight \( t \) is given as 1 second. The vertical motion can be analyzed using the equation of motion:

\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \( s \) is the vertical displacement (final position - initial position)
- \( u \) is the initial vertical velocity component
- \( a \) is the acceleration (which is \(-g\), with \( g = 10 \, \text{m/s}^2 \))
- \( t \) is the time of flight

### Step 4: Calculate the Vertical Displacement
The particle starts at (0, 1) and falls to the ground (y = 0). Therefore, the vertical displacement \( s \) is:

\[ s = 0 - 1 = -1 \, \text{m} \]

### Step 5: Substitute Values into the Equation
Substituting the known values into the equation:

\[ -1 = u_y \cdot 1 + \frac{1}{2} (-10) \cdot (1^2) \]

This simplifies to:

\[ -1 = u_y - 5 \]

Rearranging gives:

\[ u_y = 4 \, \text{m/s} \]

### Step 6: Find the Horizontal Component
Now, we need to find the horizontal component of the initial velocity \( u_x \). We know:

\[ \tan \theta = \frac{u_y}{u_x} \]

From the slope, we have:

\[ \tan \theta = 2 \implies \frac{4}{u_x} = 2 \]

Solving for \( u_x \):

\[ u_x = \frac{4}{2} = 2 \, \text{m/s} \]

### Step 7: Calculate the X-coordinate
The x-coordinate where the particle falls can be calculated using:

\[ x = u_x \cdot t \]

Substituting the values:

\[ x = 2 \cdot 1 = 2 \, \text{m} \]

### Conclusion
The x-coordinate where the particle fell is:

\[ \boxed{2} \]

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