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MMH After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 51.0 m horizontally from the end of the ramp. His velocity, just before landing, is 23.0 m/s and points in a diretion 43.0-below the horizontal. Neglecting air resistance and any lift he experiences while airborne, find his initial velocity (magnitude and direction) when he left the end of the ramp. Express the direction as an angle relative to the horizontal.

Short Answer

Expert verified
Initial velocity is 21.97 m/s at 39.3° above the horizontal.

Step by step solution

01

Analyze Final Velocity Components

The skier's velocity just before landing is given as 23.0 m/s at an angle of 43.0° below the horizontal. Break this velocity into horizontal and vertical components using trigonometric functions. Horizontal component \( v_{x_f} = v_f \cdot \cos(43.0^\circ) \) and vertical component \( v_{y_f} = v_f \cdot \sin(43.0^\circ) \). Calculating these gives:\[v_{x_f} = 23.0 \times \cos(43.0^\circ) \approx 16.80 \, \text{m/s}\]\[v_{y_f} = 23.0 \times \sin(43.0^\circ) \approx 15.69 \, \text{m/s (downward)}\]
02

Use Horizontal Motion to Find Time of Flight

Since horizontal acceleration is zero, the horizontal velocity is constant. Use the formula:\[x = v_{x} \cdot t\]where \( x = 51.0 \) m and \( v_{x} = 16.80 \) m/s. Solving for time \( t \):\[ t = \frac{x}{v_{x}} = \frac{51.0}{16.80} \approx 3.04 \, \text{seconds}\]
03

Calculate Initial Vertical Velocity

Use the vertical motion equation considering downward as positive:\[v_{y_f} = v_{y_i} + gt\]where \( g \approx 9.81 \, \text{m/s}^2 \). To account for the direction of initial velocity being upward (opposit of final direction), so take \( v_{y_f} = 15.69 \, \text{m/s} = -v_{y_i} + 9.81 \cdot 3.04 \). Rearranging gives: \[v_{y_i} = -15.69 + 29.82 \approx 14.13 \, \text{m/s (upward)}\]
04

Calculate Initial Horizontal Velocity

Since horizontal velocity is not changing, the initial horizontal velocity is the same as the final horizontal velocity:\[v_{x_i} = 16.80 \, \text{m/s}\]
05

Determine Initial Velocity Magnitude and Direction

Use the Pythagorean theorem to find the magnitude of the initial velocity:\[v_i = \sqrt{v_{x_i}^2 + v_{y_i}^2}\]Substituting the values, we get:\[v_i = \sqrt{(16.80)^2 + (14.13)^2} \approx 21.97 \, \text{m/s}\]For the direction, use the tangent function:\[\theta_i = \tan^{-1}\left(\frac{v_{y_i}}{v_{x_i}}\right) = \tan^{-1}\left(\frac{14.13}{16.80}\right) \approx 39.3^\circ \, \text{above the horizontal}\]
06

Final Answer

The magnitude of the initial velocity is approximately 21.97 m/s and the direction is 39.3° above the horizontal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Velocity Calculation
In projectile motion, the initial velocity is an essential aspect to determine the trajectory and behavior of the motion. Calculating the initial velocity involves determining both its magnitude and direction right at the onset of motion. For the ski jumper, we first need to identify his velocity components just before landing, which includes breaking down the velocity into horizontal and vertical components. With the initial vertical and horizontal velocities, we can later combine them using the Pythagorean theorem to find the initial total velocity's magnitude. This involves the formula:\[ v_i = \sqrt{(v_{x_i})^2 + (v_{y_i})^2} \]Using this process, students can clearly understand how these velocities are interrelated, enabling them to solve for the initial velocity thoroughly.
Horizontal and Vertical Components
The horizontal and vertical components of velocity are fundamental in determining the motion's characteristics. For a projectile, such as the ski jumper, the velocity is split into two perpendicular components:
  • Horizontal Component: This remains constant during the flight as there is no horizontal acceleration. It's calculated using the cosine of the angle below horizontal given before impact.
  • Vertical Component: This changes due to the influence of gravity. Initially, it can be calculated by adjusting the final vertical velocity using the equation of motion.
For example, if the horizontal velocity before landing is 16.8 m/s, it will also be the initial horizontal velocity. The principles of trigonometry help decompose the final velocity, like using the sine and cosine functions for angles. Thus, the separation into horizontal and vertical components provides a clear path to understanding the projectile’s trajectory.
Time of Flight
The time of flight is the total time the projectile remains in the air. It's crucial for predicting where and when the projectile will land. For the ski jumper, the horizontal motion equation helps determine the time of flight because:
  • The horizontal distance is known (51.0 m).
  • The horizontal velocity is constant at 16.8 m/s.
This leads to the formula: \[ t = \frac{x}{v_{x}} = \frac{51.0}{16.80} \approx 3.04 \, \text{seconds} \]Thus, in the absence of air resistance, the constant horizontal velocity allows for a straightforward calculation of the time of flight, which plays an integral role in understanding the motion fully.
Angle of Projection
Understanding the angle of projection is vital as it affects the trajectory and range of the projectile. The angle refers to the direction of the initial velocity relative to the horizontal. For the ski jumper, the initial angle of projection is calculated using the tangent function, which relates the vertical and horizontal components:\[ \theta_i = \tan^{-1}\left(\frac{v_{y_i}}{v_{x_i}}\right) \]For example, with the jumper, an upward angle is determined from the horizontal. This angle is 39.3° above the horizontal, indicating the skier's initial launch direction. Understanding this angle is crucial because it determines the initial path of the projectile, influencing how far and high it will travel. Grasping this concept allows students to predict trajectories in other projectile motion scenarios too.

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