Projectile Motion - Skier & Ramp

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Homework Help Overview

The discussion revolves around a ski jumper's projectile motion after launching off a ramp at a speed of 119.0 km/hr, with the ground sloping downward at a 45-degree angle. The original poster seeks to determine the distance down the slope where the skier lands, assuming free-fall motion post-launch.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the landing distance using a kinematic equation but questions their approach after receiving feedback. Some participants suggest revisiting the SUVAT equations and solving from first principles. Others propose using the angle of the slope to derive the distance using trigonometric relationships.

Discussion Status

Participants are actively exploring different interpretations of the problem, with some providing guidance on the correct application of physics principles. There is an acknowledgment of the need for the original poster to engage with the problem-solving process themselves.

Contextual Notes

There is a noted confusion regarding the application of horizontal motion equations in the context of a slope, as well as the role of gravity acting vertically rather than parallel to the slope.

Denize
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Homework Statement


A ski jumper acquires a speed of 119.0 km/hr by racing down a steep hill. He then lifts off into the air from a horizontal ramp. Beyond this ramp, the ground slopes downward at an angle of θ = 45 degrees.

Homework Equations


Assuming the skier is in free-fall motion after he leaves the ramp, at what distance d down the slope does the skier land?

The Attempt at a Solution


I converted 119.0 km/hr to m/s, so 33.055 m/s

I then try to find the horizontal range: 0=33.055^2 + 2*9.8*X and found X=55.74m
55.74cos45=78.83m down the ramp.

What have I done wrong?
 
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Denize said:
the horizontal range: 0=33.055^2 + 2*9.8*X
That formula is for taking off and landing at the same height.
Go back to the SUVAT equations and solve from first principles.
 
With respect to the slope which is angled at 45 dgree, the ski jumper is jumping off the ramp with a velocity u=(119x5/18)m/s at angle of 45 degree!
Thus, the horizontal range he covers on the slope, d=(u^2xsin(2θ)/g).
Correct me if I am wrong,or you can ask for further doubts or queries.
 
Avimanyu Ray said:
With respect to the slope which is angled at 45 dgree, the ski jumper is jumping off the ramp with a velocity u=(119x5/18)m/s at angle of 45 degree!
Thus, the horizontal range he covers on the slope, d=(u^2xsin(2θ)/g).
Correct me if I am wrong,or you can ask for further doubts or queries.
You are wrong. Horizontal means horizontal - it doesn't mean parallel to the slope.
Gravity continues to act vertically, not orthogonally to the slope.
 
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haruspex said:
You are wrong. Horizontal means horizontal - it doesn't mean parallel to the slope.
Gravity continues to act vertically, not orthogonally to the slope.
Thanks for the glitch. :oldsmile:
Ok, suppose 'd' is the hypotenuse of an isosceles triangle of angle 45 degree and its equal sides=dcos45 or (dsin45 whatever). Then time taken for the skii jumper to cover 'd' is t=(2xh/g)½ = (2xdcos45/g)½.
Now, R=dsin45=u x t= (u^2 x 2 x dcos45/g)½
solving out the above equation will give the value of 'd'.
 
Avimanyu Ray said:
solving out the above equation will give the value of 'd'.
Yes it's correct now but you should let the original poster figure it out for themselves so they can learn too :oldsmile:
 
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