When an object is shot from the bottom of a long smooth inclined plane kept at an angle `60^(@)` with horizontal, it can travel a distance `x_(1)` along the plane. But when the inclination is decreased to `30^(@)` and the same object is shot with the same velocity, it can travel `x_(2)` distance. Then `x_(1):x_(2)` will be :

Options

A. `1:2sqrt(3)`
B. `1:sqrt(2)`
C. `sqrt(2):1`
D. `1:sqrt(3)`

Explanation

To solve the problem, we need to find the ratio \( x_1 : x_2 \) where \( x_1 \) is the distance traveled by the object on a smooth inclined plane at an angle of \( 60^\circ \) and \( x_2 \) is the distance traveled at an angle of \( 30^\circ \).

### Step 1: Understand the forces acting on the object
When the object is shot up the incline, the forces acting on it are:
- Gravitational force \( mg \) acting downwards.
- Normal force \( N \) acting perpendicular to the inclined plane.

The component of gravitational force acting down the incline is \( mg \sin \theta \).

### Step 2: Determine the acceleration of the object
The acceleration \( a \) of the object moving up the incline can be expressed as:
\[ a = -g \sin \theta \]
This is negative because the gravitational force is acting opposite to the direction of motion.

### Step 3: Use the kinematic equation
Using the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) is the final velocity (0 when the object comes to rest),
- \( u \) is the initial velocity (same in both cases),
- \( a \) is the acceleration,
- \( s \) is the distance traveled.

Rearranging gives:
\[ 0 = u^2 + 2(-g \sin \theta)s \]
This simplifies to:
\[ u^2 = 2g \sin \theta \cdot s \]
Thus, we can express the distance \( s \) (which is \( x_1 \) or \( x_2 \)) as:
\[ s = \frac{u^2}{2g \sin \theta} \]

### Step 4: Calculate \( x_1 \) and \( x_2 \)
For \( x_1 \) (when \( \theta = 60^\circ \)):
\[ x_1 = \frac{u^2}{2g \sin 60^\circ} \]
For \( x_2 \) (when \( \theta = 30^\circ \)):
\[ x_2 = \frac{u^2}{2g \sin 30^\circ} \]

### Step 5: Find the ratio \( x_1 : x_2 \)
Now, we can find the ratio:
\[ \frac{x_1}{x_2} = \frac{\frac{u^2}{2g \sin 60^\circ}}{\frac{u^2}{2g \sin 30^\circ}} = \frac{\sin 30^\circ}{\sin 60^\circ} \]
Substituting the values of sine:
\[ \sin 30^\circ = \frac{1}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2} \]
Thus:
\[ \frac{x_1}{x_2} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \]

### Final Answer
Therefore, the ratio \( x_1 : x_2 \) is:
\[ x_1 : x_2 = 1 : \sqrt{3} \]