How Far Will a Skier Land After a Flying Ski Jump?

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In a flying ski jump, a skier reaches a speed of 110 km/h before launching off a horizontal ramp, with the ground sloping downward at a 45-degree angle. The calculations indicate that the skier's horizontal and vertical distances are equal due to the slope's angle, leading to a projectile motion analysis. The initial velocity components are determined, with horizontal velocity calculated as 30.6 m/s and vertical velocity as 0. The time to impact is derived from the relationship between horizontal and vertical distances. The calculated landing distance does not match actual jump distances of up to 165m, likely due to factors like air resistance and jump technique not being considered.
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Homework Statement


In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a horizontal ramp. Beyond this ramp, the ground slopes downward at an angle of 45 degrees. (a) Assuming the skier is in a free-fall motion after he leaves the ramp, at what distance down the slope will he land? (b) In actual jumps, skiers reach distances of up to 165m. Why does this not agree with the result you obtained in part (a)?


Homework Equations


vy= voy-gt
y-yo= voyt + 1/2(ayt^2)


The Attempt at a Solution

 
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Since the slope is 45 degrees, horizontal distance and the vertical fall are equal.. Find those quantities and equate them.
 
I thought this looked like a projectile question. The question didn't mention a take of angle so looking at what was given I think that angle is horizontal so its 0 degrees.

You need to vectorise everything two because this is a two dimensional question.
The velocity of take off first- Vx= V Cos theta
= 30.6 m/s
and Vy= V Sin theta
= 0
Then substitute those values into the vector for distance in the x and y plane.
The skier will hit the slopes when distancex/distance y= Tan -45 degrees and you'll find t the time it took to hit the hill.
To find the downhill distance draw a triangle with x distance on the on the horizontal and y distance on the vertical and 50 degrees for their facing angle so x distance /Cos 50 should get you that distance.
I came up with a huge figure. I get friction etc wasn't taken into account,
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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