A skier is gliding along at 3.2 m/s
 on horizontal, frictionless snow. He suddenly starts down a 10∘
 incline. His speed at the bottom is 20 m/s
 . What is the length of the incline?

To solve for the length of the incline, we can use the concept of energy conservation, as there is no friction involved and only gravitational potential energy is being converted into kinetic energy.

### Given dataTo solve for the length of the incline, we can use the concept of energy conservation, as there is no friction involved and only gravitational potential energy is being converted into kinetic energy.

### Given 
- Initial speed \( v_i = 3.2 \, \text{m/s} \)
- Final speed \( v_f = 20 \, \text{m/s} \)
- Incline angle \( \theta = 10^\circ \)

We will use the principle of **conservation of mechanical energy**. The skier's initial energy consists of kinetic energy at the top of the incline, and at the bottom, it's fully converted to kinetic energy.

### Step 1: Kinetic Energy at the Top and Bottom
The skier has only kinetic energy both at the top and the bottom. Since the snow is frictionless, potential energy is lost as the skier descends the incline and converted into kinetic energy.

- Initial kinetic energy: 
\[
KE_i = \frac{1}{2} m v_i^2
\]
- Final kinetic energy: 
\[
KE_f = \frac{1}{2} m v_f^2
\]

### Step 2: Energy Conservation
The difference in kinetic energy is equal to the loss in potential energy:

\[
KE_f - KE_i = mgh
\]

Where \( h \) is the vertical height of the incline and can be related to the incline's length \( L \) and angle \( \theta \) as:

\[
h = L \sin \theta
\]

Thus, the equation becomes:

\[
\frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 = mgL \sin \theta
\]

### Step 3: Simplifying and Solving for \( L \)
The mass \( m \) cancels out from all terms, and we can solve for the length \( L \):

\[
\frac{1}{2} (v_f^2 - v_i^2) = g L \sin \theta
\]

Rearranging for \( L \):

\[
L = \frac{v_f^2 - v_i^2}{2g \sin \theta}
\]

### Step 4: Substituting the Values
- \( v_f = 20 \, \text{m/s} \)
- \( v_i = 3.2 \, \text{m/s} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( \theta = 10^\circ \)

Now, calculate \( L \):

\[
L = \frac{(20)^2 - (3.2)^2}{2 \cdot 9.8 \cdot \sin(10^\circ)}
\]

Let me compute the result.The length of the incline is approximately \( 114.52 \, \text{meters} \).

Would you like more details or have any other questions?

Here are 5 related questions for further exploration:
1. How would the result change if the angle of the incline increased to \( 15^\circ \)?
2. What would happen if friction were introduced? How would it affect the length of the incline?
3. How can we calculate the skier's time to reach the bottom of the incline?
4. What is the skier's acceleration while descending the incline?
5. How would this scenario differ if the skier started from rest at the top?

**Tip:** When solving incline problems, always relate height to the length of the incline using trigonometric functions like sine or cosine, based on the problem's geometry.

A skier is gliding along at 3.2 m/s on horizontal, frictionless snow. He suddenly starts down a 10° incline. His speed at the bottom is 20 m/s. What is the length of the incline?

Learn how to calculate the length of an incline based on a skier's initial and final speeds, using energy conservation and trigonometry. This detailed solution explains key concepts such as kinetic energy and incline geometry, suitable for Grades 10-12.

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