A body is slipping from an inclined plane of height h and length l . If the angle of inclination is `theta` , the time taken by the body to come from the top to the bottom of this inclined plane is

Options

A. `sqrt((2h)/(g))`
B. `sqrt((2l)/(g))`
C. `(1)/(sin theta)sqrt((2h)/(g))`
D. `sin theta sqrt((2h)/(g))`

Explanation

To solve the problem of finding the time taken by a body to slide down an inclined plane of height \( h \) and length \( l \) at an angle \( \theta \), we can follow these steps:

### Step 1: Identify the forces acting on the body
The body is subjected to gravitational force \( mg \). The component of this force acting down the incline is \( mg \sin \theta \), while the component acting perpendicular to the incline is \( mg \cos \theta \).

### Step 2: Determine the acceleration of the body
Using Newton's second law, the net force acting down the incline is:
\[ F = mg \sin \theta \]
Thus, the acceleration \( a \) of the body can be expressed as:
\[ a = \frac{F}{m} = g \sin \theta \]

### Step 3: Use the kinematic equation
We can use the kinematic equation for motion along the incline:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( s \) is the distance traveled along the incline (which is \( l \)), \( u \) is the initial velocity (which is 0), and \( a \) is the acceleration we found earlier. Plugging in the values, we get:
\[ l = 0 \cdot t + \frac{1}{2} (g \sin \theta) t^2 \]
This simplifies to:
\[ l = \frac{1}{2} g \sin \theta \cdot t^2 \]

### Step 4: Solve for time \( t \)
Rearranging the equation to solve for \( t^2 \):
\[ t^2 = \frac{2l}{g \sin \theta} \]
Taking the square root gives:
\[ t = \sqrt{\frac{2l}{g \sin \theta}} \]

### Step 5: Relate \( \sin \theta \) to height \( h \) and length \( l \)
From the geometry of the inclined plane, we know:
\[ \sin \theta = \frac{h}{l} \]
Substituting this into the equation for \( t \):
\[ t = \sqrt{\frac{2l}{g \cdot \frac{h}{l}}} = \sqrt{\frac{2l^2}{gh}} \]

### Final Result
Thus, the time taken by the body to slide down the inclined plane is:
\[ t = \sqrt{\frac{2l^2}{gh}} \]