To solve this problem, we can use some basic physics involving projectile motion and geometry.
Part 1: Distance along the incline before landing
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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.
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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.
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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: \[ y = \frac{1}{2} g t^2 = x \tan(37^\circ) \]
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Finding the relationship between \( x \) and \( t \): The horizontal distance the jumper travels before hitting the slope is: \[ x_{horizontal} = v_0 t \]
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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:
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Horizontal distance: \[ v_0 t = x \cos(37^\circ) \] \[ t = \frac{x \cos(37^\circ)}{v_0} \]
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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:
- Distance along the incline: \( x \approx 174.50 , \text{m} \)
- Time in air: \( t \approx 5.17 , \text{s} \)
- Impact angle: \( \phi \approx 62.26^\circ \)