To analyze the ski jumper's motion as she leaves the slope horizontally, we'll define the situation in terms of position, velocity, and time. The key points to consider are the horizontal and vertical motions of the jumper separately, taking into consideration gravity’s effect on her descent.
Initial Conditions
- Initial horizontal velocity (\(v_{0x}\)): The horizontal speed when leaving the ski track is \(27 \text{ m/s}\).
- Initial vertical velocity (\(v_{0y}\)): Since the jumper leaves horizontally, the initial vertical velocity is \(0 \text{ m/s}\).
Motion in the Vertical Direction
The only force acting on the jumper in the vertical direction is gravity. The vertical motion can be described by the following kinematic equation:
\[ y = v_{0y} t + \frac{1}{2} g t^2 \]
Since \(v_{0y} = 0\), this simplifies to:
\[ y = \frac{1}{2} g t^2 \]
Motion in the Horizontal Direction
In the horizontal direction, there are no forces acting on the jumper (assuming air resistance is negligible). Therefore, the horizontal position can be described by:
\[ x = v_{0x} t \]
Finding the Time Until the Jumper Hits the Incline
Now we need to find the point where the jumper intersects the landing incline, which has a slope of \(37^\circ\). The equation of the incline can be described using the tangent of the angle:
\[ y = x \tan(\theta) \]
Substituting \(\theta = 37^\circ\):
\[ y = x \tan(37^\circ) \]
We need to express \(y\) in terms of \(x\) using the time variable \(t\). From the horizontal motion:
\[ x = v_{0x} t \implies t = \frac{x}{v_{0x}} \]
Substituting this into the equation for vertical motion:
\[ y = \frac{1}{2} g t^2 = \frac{1}{2} g \left(\frac{x}{v_{0x}}\right)^2 \]
Now we can set the two expressions for \(y\) equal to each other:
\[ \frac{1}{2} g \left(\frac{x}{v_{0x}}\right)^2 = x \tan(37^\circ) \]
Solving for \(x\)
Rearranging gives:
\[ \frac{1}{2} g \frac{x^2}{v_{0x}^2} = x \tan(37^\circ) \]
Assuming \(x \neq 0\), we can divide both sides by \(x\):
\[ \frac{1}{2} g \frac{x}{v_{0x}^2} = \tan(37^\circ) \]
Now solve for \(x\):
\[ x = \frac{2 v_{0x}^2 \tan(37^\circ)}{g} \]
Plugging in Values
- \(g = 9.8 , \text{m/s}^2\)
- \(v_{0x} = 27 , \text{m/s}\)
- \(\tan(37^\circ) \approx 0.7536\)
Now substituting these values provides:
\[ x = \frac{2 \cdot (27)^2 \cdot 0.7536}{9.8} \]
Calculating:
\[ x = \frac{2 \cdot 729 \cdot 0.7536}{9.8} \approx \frac{1095.12}{9.8} \approx 111.8 , \text{m} \]
Finding the Time to Fall
Now using the value of \(x\) to find \(t\):
\[ t = \frac{x}{v_{0x}} = \frac{111.8 , \text{m}}{27 , \text{m/s}} \approx 4.14 , \text{s} \]
Summary
- The horizontal distance traveled before landing is approximately 111.8 meters.
- The time of flight until the jumper hits the incline is approximately 4.14 seconds.
This method separates the motion into horizontal and vertical components, allowing us to calculate when the jumper hits the incline safely.