A spherical ball of mass 2 kg is stationalry at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then cilimbs up another hill of height 30 m and finally rolls down on a smooth surface to a horizontal base at a height of 20 m abvoe the ground. What is the final velocity attained by the ball? `(g=10m//s^(2))`

Options

A. `10sqrt(30)m//s`
B. `10m//s`
C. `20m//s`
D. `40m//s`

Explanation

To find the final velocity attained by the ball after it rolls down the hills and smooth surfaces, we can use the principle of conservation of energy. The potential energy at the top of the hills will be converted into kinetic energy as the ball rolls down.

### Step-by-step Solution:

1. **Calculate the initial potential energy (PE_initial)** at the top of the first hill:
\[ PE_{\text{initial}} = mgh \]
where \( m = 2 \, \text{kg} \), \( g = 10 \, \text{m/s}^2 \), and \( h = 100 \, \text{m} \).
\[ PE_{\text{initial}} = 2 \times 10 \times 100 = 2000 \, \text{J} \]

2. **Calculate the potential energy (PE) at the top of the second hill (30 m)**:
\[ PE_{\text{second hill}} = mgh = 2 \times 10 \times 30 = 600 \, \text{J} \]

3. **Calculate the potential energy (PE) at the height of 20 m**:
\[ PE_{\text{final}} = mgh = 2 \times 10 \times 20 = 400 \, \text{J} \]

4. **Calculate the total mechanical energy at the start and at the end**:
- Total mechanical energy at the start (top of the first hill):
\[ E_{\text{initial}} = PE_{\text{initial}} = 2000 \, \text{J} \]
- Total mechanical energy at the end (at height 20 m):
\[ E_{\text{final}} = PE_{\text{final}} + KE_{\text{final}} \]
where \( KE_{\text{final}} \) is the kinetic energy at the height of 20 m.

5. **Set the initial energy equal to the final energy**:
\[ E_{\text{initial}} = E_{\text{final}} \]
\[ 2000 = 400 + KE_{\text{final}} \]
\[ KE_{\text{final}} = 2000 - 400 = 1600 \, \text{J} \]

6. **Relate kinetic energy to velocity**:
\[ KE = \frac{1}{2} mv^2 \]
\[ 1600 = \frac{1}{2} \times 2 \times v^2 \]
\[ 1600 = v^2 \]
\[ v = \sqrt{1600} = 40 \, \text{m/s} \]

### Final Answer:
The final velocity attained by the ball is \( 40 \, \text{m/s} \).