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: \[ \boxed{140.5} , \text{m} \]