A ball of mass `0.2 kg` and radius `0.5 m` starting from rest rolls down a `30^(circ)` inclined plane. Find the time in second it would take to cover `7 m`.

Answer

To solve the problem of a ball rolling down a 30-degree inclined plane, we will follow these steps:

### Step 1: Identify the parameters
- Mass of the ball, \( m = 0.2 \, \text{kg} \)
- Radius of the ball, \( r = 0.5 \, \text{m} \)
- Angle of inclination, \( \theta = 30^\circ \)
- Distance to cover, \( s = 7 \, \text{m} \)
- Initial velocity, \( u = 0 \, \text{m/s} \) (the ball starts from rest)

### Step 2: Calculate the acceleration of the ball
For a solid sphere rolling down an incline, the acceleration \( a \) can be calculated using the formula:
\[ a = \frac{g \sin \theta}{1 + k^2} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)) and \( k^2 \) is the radius of gyration. For a solid sphere, \( k^2 = \frac{2}{5} \).

Substituting the values:
\[ a = \frac{9.8 \sin(30^\circ)}{1 + \frac{2}{5}} \]
Since \( \sin(30^\circ) = 0.5 \):
\[ a = \frac{9.8 \times 0.5}{1 + 0.4} = \frac{4.9}{1.4} = 3.5 \, \text{m/s}^2 \]

### Step 3: Use the kinematic equation to find time
We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \):
\[ s = \frac{1}{2} a t^2 \]
Substituting \( s = 7 \, \text{m} \) and \( a = 3.5 \, \text{m/s}^2 \):
\[ 7 = \frac{1}{2} \times 3.5 \times t^2 \]
\[ 7 = 1.75 t^2 \]
\[ t^2 = \frac{7}{1.75} = 4 \]
\[ t = \sqrt{4} = 2 \, \text{s} \]

### Final Answer
The time it would take for the ball to cover 7 meters is \( t = 2 \, \text{seconds} \).

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