A skier leaves the top of a slope with an initial speed of 5.0 m/s. Her speed at the bottom of the slope is 13 m/s. What is the height of the slope ?

Options

A. 1.1m
B. 4.6m
C. 6.4m
D. 7.3m

Answer

To find the height of the slope that the skier descends, we can use the principle of conservation of energy, specifically the work-energy theorem. This theorem states that the work done by the forces acting on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial speed of the skier, \( v_i = 5.0 \, \text{m/s} \)
- Final speed of the skier, \( v_f = 13.0 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

2. **Write the Work-Energy Theorem:**
The work done by gravity is equal to the change in kinetic energy:
\[ W = \Delta KE = KE_f - KE_i \]
where \( KE = \frac{1}{2} mv^2 \).

3. **Calculate the Initial and Final Kinetic Energies:**
- Initial kinetic energy, \( KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} m (5.0)^2 = \frac{1}{2} m (25) = 12.5m \)
- Final kinetic energy, \( KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (13.0)^2 = \frac{1}{2} m (169) = 84.5m \)

4. **Calculate the Change in Kinetic Energy:**
\[ \Delta KE = KE_f - KE_i = 84.5m - 12.5m = 72m \]

5. **Set Up the Equation for Work Done by Gravity:**
The work done by gravity can also be expressed as:
\[ W = mgh \]
where \( h \) is the height of the slope.

6. **Equate the Work Done to the Change in Kinetic Energy:**
\[ mgh = 72m \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = 72 \]

7. **Solve for Height \( h \):**
\[ h = \frac{72}{g} = \frac{72}{9.8} \]
\[ h \approx 7.35 \, \text{m} \]

### Final Answer:
The height of the slope is approximately \( 7.35 \, \text{m} \).