Question

Part 1: A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 27 m/s as in the figure. The landing incline below her falls off with a slope of θ = 37 degrees. The acceleration of gravity is 9.8 m/s^2. Calculate the distance she travels along the incline before landing. Answer in units of m.

Part 2: Determine how long the ski jumper is airborne. Answer in units of s.

Part 3: What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of degrees.

Answer 1

To solve this problem, we can use some basic physics involving projectile motion and geometry.

### Part 1: Distance along the incline before landing

1. **Identify the components of motion**: The ski jumper leaves the ramp horizontally with a speed \( v_0 = 27 \, \text{m/s} \), and we assume \( t = 0 \) when she leaves the slope. The vertical motion is governed by gravity.

2. **Calculate the time of flight**:
The vertical position during free fall can be described by the equation:
\[
y = \frac{1}{2} g t^2
\]
where \( g = 9.8 \, \text{m/s}^2 \). Since the ski jumper will hit the inclined plane at a vertical height \( y \) defined by how far she falls before hitting that incline.

3. **Relationship with the incline**:
The incline at angle \( \theta = 37^\circ \) can be described by:
\[
y = x \tan(\theta)
\]
where \( x \) is the distance she travels along the incline. Combining both equations:
\[
y = \frac{1}{2} g t^2 = x \tan(37^\circ)
\]

4. **Finding the relationship between \( x \) and \( t \)**: The horizontal distance the jumper travels before hitting the slope is:
\[
x_{horizontal} = v_0 t
\]

5. **Use the geometric relationship to solve for \( x \)**:
The horizontal distance \( x_{horizontal} \) can also be expressed in terms of \( x \) as:
\[
x_{horizontal} = x \cos(37^\circ)
\]
Hence, we can set \( v_0 t = x \cos(37^\circ) \).

Now substituting \( y = x \tan(37^\circ) \):
\( x \tan(37^\circ) = \frac{1}{2} g t^2 \).

### Equations Set up:
1. Horizontal distance:
\[
v_0 t = x \cos(37^\circ)
\]
\[
t = \frac{x \cos(37^\circ)}{v_0}
\]

2. Vertical distance (substituting \( t \)):
\[
x \tan(37^\circ) = \frac{1}{2} g \left(\frac{x \cos(37^\circ)}{v_0}\right)^2
\]

Rearranging to isolate \( x \):
\[
x \tan(37^\circ) = \frac{1}{2} g \frac{x^2 \cos^2(37^\circ)}{v_0^2}
\]
Cancel \( x \) (assuming \( x \neq 0 \)):
\[
\tan(37^\circ) = \frac{1}{2} g \frac{x \cos^2(37^\circ)}{v_0^2}
\]

Thus,
\[
x = \frac{2 v_0^2 \tan(37^\circ)}{g \cos^2(37^\circ)}
\]

Plug in the known values:
- \( v_0 = 27 \, \text{m/s} \),
- \( g = 9.8 \, \text{m/s}^2 \),
- \( \tan(37^\circ) \approx 0.7536 \),
- \( \cos(37^\circ) \approx 0.7986 \).

Therefore:
\[
x = \frac{2 (27)^2 (0.7536)}{9.8 (0.7986)^2}
= \frac{2 \times 729 \times 0.7536}{9.8 \times 0.6386}
= \frac{1095.1304}{6.2654} \approx 174.50 \, \text{m}
\]

### Part 2: Time in the air
Using \( t \) from the horizontal velocity calculations:
\[
t = \frac{x \cos(37^\circ)}{v_0}
= \frac{174.50 \cdot 0.7986}{27} \approx \frac{139.63}{27} \approx 5.17 \, s
\]

### Part 3: Angle of impact φ
The vertical velocity component \( v_y \) when landing is given as:
\[
v_y = g t \ = 9.8 \cdot 5.17 \approx 50.75 \, \text{m/s}
\]

The horizontal \( v_x = v_0 = 27 \, \text{m/s} \).

Using \( \tan(\phi) = \frac{v_y}{v_x} \):
\[
\tan(\phi) = \frac{50.75}{27}
\]
Calculating \( \phi \):
\[
\phi = \arctan\left(\frac{50.75}{27}\right) \approx \arctan(1.88) \approx 62.26^\circ
\]

Thus, we summarize the answers:
1. **Distance along the incline:** \( x \approx 174.50 \, \text{m} \)
2. **Time in air:** \( t \approx 5.17 \, \text{s} \)
3. **Impact angle:** \( \phi \approx 62.26^\circ \)