To solve the problem, we can break it into three parts as requested.
Part 1: Distance Traveled Along the Incline
Given:
- Initial horizontal velocity, \( v_0 = 27 , \text{m/s} \)
- Angle of the slope, \( \theta = 37^\circ \)
- Acceleration due to gravity, \( g = 9.8 , \text{m/s}^2 \)
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Finding time of flight:
The ski jumper is in projectile motion, leaving the ramp horizontally. The vertical motion can be analyzed separately. The vertical displacement \( y \) can be described by the equation: \[ y = \frac{1}{2} g t^2 \]During this time, the horizontal distance \( x \) traveled is: \[ x = v_0 t \]
Since the landing incline has a slope of \( \theta \), we can express the relation between horizontal distance \( x \) and vertical distance \( y \): \[ y = x \tan(\theta) \]
Substitute \( x = v_0 t \): \[ y = v_0 t \tan(\theta) \]
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Equate the equations: Now we can equate the two expressions for \( y \): \[ \frac{1}{2} g t^2 = v_0 t \tan(\theta) \]
Simplifying, we can cancel \( t \) from both sides (assuming \( t \neq 0 \)): \[ \frac{1}{2} g t = v_0 \tan(\theta) \]
Solving for \( t \): \[ t = \frac{2 v_0 \tan(\theta)}{g} \]
Substitute the known values: \[ t = \frac{2 \times 27 , \text{m/s} \times \tan(37^\circ)}{9.8 , \text{m/s}^2} \]
First, calculate \( \tan(37^\circ) \): \[ \tan(37^\circ) \approx 0.7536 \]
Now substituting: \[ t \approx \frac{2 \times 27 \times 0.7536}{9.8} \approx \frac{40.3072}{9.8} \approx 4.116 , \text{s} \]
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Calculate the distance along the incline: The distance \( s \) along the incline can be calculated using: \[ s = x/\cos(\theta) \]
Where \( x = v_0 t \): \[ x = 27 \times 4.116 \approx 111.132 , \text{m} \] Thus, the distance along the incline: \[ s = \frac{x}{\cos(37^\circ)} = \frac{111.132}{0.7986} \approx 139.2 , \text{m} \]
Part 2: Time of Flight
From Part 1, we found the time of flight to be: \[ t \approx 4.116 , \text{s} \]
Part 3: Relative Angle of Impact
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Finding the angle of impact \( \phi \): The impact angle can be found using the path of the ski jumper. Just before impacting the slope, the vertical velocity \( v_y \) is given by: \[ v_y = g t \] Substitute \( t \): \[ v_y = 9.8 \times 4.116 \approx 40.4 , \text{m/s} \]
The horizontal velocity \( v_x \) remains constant at: \[ v_x = 27 , \text{m/s} \]
The angle of impact \( \phi \) can then be found using: \[ \tan(\phi) = \frac{v_y}{v_x} \] Substitute the values: \[ \tan(\phi) = \frac{40.4}{27} \]
Calculate \( \phi \): \[ \phi \approx \tan^{-1}\left(\frac{40.4}{27}\right) \approx \tan^{-1}(1.494) \approx 56.3^\circ \]
Summary of Answers:
- Distance along the incline: \( s \approx 140.5 , \text{m} \)
- Time of flight: \( t \approx 4.116 , \text{s} \)
- Angle of impact: \( \phi \approx 56.3^\circ \)