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. (3,0)
B. (4,0)
C. (2,0)
D. (2`sqrt5`,0)

Explanation

To solve the problem step by step, we will follow the kinematic equations of motion and the geometry of the projectile motion.

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

### Step 2: Determine the angle of projection
The angle of projection can be determined using the coordinates of the two points. The vertical displacement (y-axis) is from 1 to 9, which is \(9 - 1 = 8\) meters, and the horizontal displacement (x-axis) is from 0 to 4, which is \(4 - 0 = 4\) meters.

Using the tangent function:
\[ \tan \theta = \frac{\text{Vertical Displacement}}{\text{Horizontal Displacement}} = \frac{8}{4} = 2 \]
Thus, \(\tan \theta = 2\).

### Step 3: Write the equation of motion in the vertical direction
The vertical motion can be described by the equation:
\[ y = y_0 + u_y t - \frac{1}{2} g t^2 \]
Where:
- \(y_0 = 1\) m (initial height)
- \(g = 10 \, \text{m/s}^2\) (acceleration due to gravity)
- \(t = 1 \, \text{s}\) (time of flight)

At the time of falling to the ground, \(y = 0\):
\[ 0 = 1 + u_y (1) - \frac{1}{2} \cdot 10 \cdot (1)^2 \]
This simplifies to:
\[ 0 = 1 + u_y - 5 \]
\[ u_y = 4 \, \text{m/s} \]

### Step 4: Relate \(u_y\) to the initial velocity \(u\) and angle \(\theta\)
The vertical component of the initial velocity is given by:
\[ u_y = u \sin \theta \]
From the previous step, we found \(u_y = 4\). Thus:
\[ 4 = u \sin \theta \]

### Step 5: Find the horizontal component of the initial velocity
Using the relationship of \(\tan \theta\):
\[ \tan \theta = \frac{u \sin \theta}{u \cos \theta} = 2 \]
This implies:
\[ \frac{u \sin \theta}{u \cos \theta} = 2 \Rightarrow \frac{4}{u \cos \theta} = 2 \]
Thus:
\[ u \cos \theta = 2 \]

### Step 6: Find the x-coordinate where it fell
The x-coordinate at the time of falling can be calculated using:
\[ x = u_x \cdot t = (u \cos \theta) \cdot t \]
Substituting \(t = 1\):
\[ x = (u \cos \theta) \cdot 1 = 2 \]

### Final Answer
The x-coordinate where the particle fell is \(2 \, \text{m}\).

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