Question

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.

Answer 1

To find the distance traveled along the incline before landing, we can break the problem down using kinematics and geometry. The jumper takes off horizontally with a speed of \( v_0 = 27 \, \text{m/s} \) and is subjected only to the acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \) in the vertical direction. The landing incline has a slope of \( \theta = 37^\circ \).

First, we will determine how long the jumper is in the air (time of flight) before they hit the incline. The path of the jumper is a projectile motion problem, with horizontal displacement and vertical displacement occurring simultaneously.

### Step 1: Determine the vertical distance fallen (using projectile motion)

The horizontal position \( x \) as a function of time \( t \) is given by:
\[
x(t) = v_0 t
\]
Since she leaves the slope horizontally, the vertical position \( y \) as a function of time can be described as:
\[
y(t) = \frac{1}{2} g t^2
\]
where \( g \) is the acceleration due to gravity.

### Step 2: Find the relationship between horizontal travel and vertical drop

The equation of the slope can be described in terms of the angle:
\[
y = x \tan(\theta)
\]

Substituting the expressions for \( x(t) \) and \( y(t) \) into the slope equation, we get:
\[
\frac{1}{2} g t^2 = v_0 t \tan(\theta)
\]

### Step 3: Rearrange to solve for \( t \)

Cancelling \( t \) from both sides (assuming \( t \neq 0 \)):
\[
\frac{1}{2} g t = v_0 \tan(\theta)
\]
Rearranging gives:
\[
t = \frac{2 v_0 \tan(\theta)}{g}
\]

### Step 4: Calculate \( \tan(37^\circ) \)

Using a calculator, we find:
\[
\tan(37^\circ) \approx 0.7536
\]

### Step 5: Substitute values to find \( t \)

Now plug in the values:
\[
t = \frac{2 \times 27 \, \text{m/s} \times 0.7536}{9.8 \, \text{m/s}^2}
\]
Calculating this gives:
\[
t = \frac{40.6848}{9.8} \approx 4.146 \, \text{s}
\]

### Step 6: Find horizontal distance \( x \)

Now we substitute \( t \) back into the horizontal displacement \( x(t) \):
\[
x = v_0 t = 27 \, \text{m/s} \times 4.146 \, \text{s} \approx 111.942 \, \text{m}
\]

### Step 7: Calculate the distance along the incline

Using the relationship between \( x \) and the distance along the incline \( d \):
\[
d = \frac{x}{\cos(\theta)}
\]
Now we calculate \( d \) where \( \cos(37^\circ) \approx 0.7986 \):
\[
d = \frac{111.942}{0.7986} \approx 140.500 \, \text{m}
\]

### Conclusion

The distance the ski jumper travels along the incline before landing is approximately:
\[
\boxed{140.5} \, \text{m}
\]