A bicyclist is traveling at a speed of 20.0 m/s as the approaches the bottom of a hill. He decides to coast up the hill and stops upon reaching the top. Neglecting friction, determine the vertical height of the hill .

Options

A. 28.5m
B. 11.2m
C. 20.4m
D. 40.8m

Answer

To solve the problem, we will use the principle of conservation of energy, specifically the work-energy theorem. The kinetic energy of the bicyclist at the bottom of the hill will be converted into gravitational potential energy at the top of the hill.

### Step-by-Step Solution:

1. **Identify the Initial Kinetic Energy (KE_initial)**:
The initial kinetic energy of the bicyclist can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m v^2 \]
where \( v = 20.0 \, \text{m/s} \) is the speed of the bicyclist.

2. **Set the Final Kinetic Energy (KE_final)**:
At the top of the hill, the bicyclist comes to a stop, so the final kinetic energy is:
\[ KE_{\text{final}} = 0 \]

3. **Identify the Change in Potential Energy (PE)**:
The potential energy at the height \( h \) is given by:
\[ PE = mgh \]
where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

4. **Apply the Work-Energy Principle**:
According to the work-energy principle, the work done on the bicyclist is equal to the change in kinetic energy:
\[ KE_{\text{final}} - KE_{\text{initial}} = -PE \]
Since \( KE_{\text{final}} = 0 \) and \( KE_{\text{initial}} = \frac{1}{2} m v^2 \), we have:
\[ 0 - \frac{1}{2} m v^2 = -mgh \]

5. **Rearranging the Equation**:
Rearranging the equation gives:
\[ mgh = \frac{1}{2} mv^2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{1}{2} v^2 \]

6. **Solve for Height (h)**:
Rearranging for \( h \):
\[ h = \frac{v^2}{2g} \]
Substituting \( v = 20.0 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \):
\[ h = \frac{(20.0)^2}{2 \times 9.8} \]
\[ h = \frac{400}{19.6} \approx 20.41 \, \text{m} \]

### Final Answer:
The vertical height of the hill is approximately \( 20.41 \, \text{m} \).

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