How Does a Skier's Launch Angle Affect Their Landing Distance?

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically analyzing how a skier's launch angle affects their landing distance on an inclined slope. The skier's initial velocity and launch angle are provided, along with the slope's angle.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations for horizontal and vertical components of velocity, questioning the correct process for determining landing distance and time of flight. There are attempts to clarify whether the focus is on distance or time.

Discussion Status

Some participants have provided equations and methods for calculating components of motion, while others are seeking clarification on the problem's requirements. There is no explicit consensus on the correct approach yet, but various interpretations and methods are being explored.

Contextual Notes

One participant notes the absence of a figure that may be crucial for understanding the problem setup. Additionally, assumptions such as negligible air resistance and the value of gravitational acceleration are mentioned.

jgroves0026
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Homework Statement


A skier leaves the ramp of a ski jump with a velocity of v = 13.0 m/s, θ = 15.0° above the horizontal, as shown in the figure. The slope where she will land is inclined downward at = 50.0°, and air resistance is negligible.


Homework Equations





The Attempt at a Solution


I thought the answer was 70.1, but I don't think that I am doing the correct process
 
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Where's the figure?
 
If you are finding the final horizontal velocity:
13cos15 = vcos50

If you are finding the final vertical velocity:
13sin15 - 9.8t = vsin 50

If you are finding the velocity, it is equal to the square root of the sum of the squares of the above two results. The direction is given by tan^-1(vx/vy).
 
Are you asking for the distance from the jump to the slope? Or the time from the jump to the slope?

Either way, you need to set up the equation [13.0 m/s*t*cos(15.0)] (i component) + [13.0 m/s*t*sin(15.0) - 1/2*g*t^2] (j component)

You can then solve for time when you divide the scalar portion of the components and set them equal to tan(-50.0) = (j component)/(i component)

Assuming g = 9.8m/s^2
t = 2.36s
 

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