A particle is projected from a point P (2,0,0)m with a velocity `10(m)/(s)` making an angle `45^@` with the horizontal. The plane of projectile motion passes through a horizontal line PQ which makes an angle of `37^@` with positive x-axis, xy plane is horizontal. The coordinates of the point where the particle will strike the line PQ is

Options

A. `(10,6,0) m`
B. `(8,6,0)m`
C. `(10,8,0) m`
D. `(6,10,0)m`

Explanation

To solve the problem, we need to find the coordinates of the point where a particle strikes the line PQ after being projected from point P (2, 0, 0) m with a velocity of 10 m/s at an angle of 45 degrees to the horizontal. The line PQ makes an angle of 37 degrees with the positive x-axis.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Initial position \( P = (2, 0, 0) \) m
- Initial velocity \( u = 10 \) m/s
- Angle of projection \( \theta = 45^\circ \)
- Angle of line PQ with the x-axis \( \phi = 37^\circ \)
- Acceleration due to gravity \( g = 10 \) m/s² (standard value)

2. **Calculate the Range of the Projectile:**
The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Substituting the values:
\[ R = \frac{10^2 \sin(90^\circ)}{10} = \frac{100 \cdot 1}{10} = 10 \text{ m} \]

3. **Determine the Coordinates of Point Q:**
The coordinates of point Q can be determined using the range and the angles involved. The x and y coordinates can be calculated using the range \( R \) and the angle \( \phi \):
- The x-coordinate change due to the range:
\[ \Delta x = R \cos(\phi) = 10 \cos(37^\circ) \]
- The y-coordinate change due to the range:
\[ \Delta y = R \sin(\phi) = 10 \sin(37^\circ) \]

Using the trigonometric values:
- \( \cos(37^\circ) = \frac{4}{5} \)
- \( \sin(37^\circ) = \frac{3}{5} \)

Now substituting these values:
\[ \Delta x = 10 \cdot \frac{4}{5} = 8 \text{ m} \]
\[ \Delta y = 10 \cdot \frac{3}{5} = 6 \text{ m} \]

4. **Calculate the Final Coordinates of Point Q:**
The coordinates of point Q can be calculated by adding the changes to the initial coordinates of point P:
\[ Q_x = 2 + \Delta x = 2 + 8 = 10 \]
\[ Q_y = 0 + \Delta y = 0 + 6 = 6 \]
\[ Q_z = 0 \text{ (since the motion is in the xy-plane)} \]

Therefore, the coordinates of point Q are:
\[ Q = (10, 6, 0) \]

### Final Answer:
The coordinates of the point where the particle will strike the line PQ are \( (10, 6, 0) \).

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