Kinematics: Final velocity components

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SUMMARY

The discussion focuses on calculating the final velocity components of a skier leaving a ramp with an initial velocity of 10.9 m/s at an angle of 13.9° above the horizontal. The slope is inclined at 48.4°, and air resistance is negligible. The participant initially calculated the time of flight as 2.965 seconds and attempted to derive the vertical and horizontal velocity components using trigonometric functions. However, confusion arose regarding the interpretation of the slope's angle, which affected the final calculations.

PREREQUISITES
  • Understanding of kinematic equations, specifically d=vit + 0.5at^2, vf=vi +at, and vf^2=Vi^2 +ad.
  • Proficiency in trigonometric functions, particularly sine and cosine for resolving velocity components.
  • Knowledge of projectile motion principles, including time of flight and maximum height calculations.
  • Familiarity with the concept of inclined planes and their impact on motion.
NEXT STEPS
  • Study the effects of slope angles on projectile motion, particularly in inclined planes.
  • Learn how to accurately resolve vectors into components using trigonometric identities.
  • Explore advanced kinematic problems involving air resistance and varying slopes.
  • Practice calculating final velocity components in different projectile scenarios using simulation tools.
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators seeking to clarify concepts related to inclined planes and velocity calculations.

huynhtn2
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Homework Statement


A skier leaves the ramp of a ski jump with a velocity of 10.9 m/s, 13.9° above the horizontal. The slope is inclined at 48.4°, and air resistance is negligible. Calculate the velocity components just before the landing. Enter the x-component (horizontal) first and then the y-component (vertical).


Homework Equations



d=vit + 0.5at^2
vf=vi +at
vf^2=Vi^2 +ad

The Attempt at a Solution



This is a second part of the question. So previously i found the total time the skier was in the air which was 2.965 seconds. I took his initial velocity and did Vsintheta to find verical velocity, and Vcostheta to find the horizontal velocity. Then i found the time where the skier reached his max height by using vf=vi +at and multiplied this time by 2.

Then i subtracted this time by the total time to see the time that he will accelerate down. I added the vertical velocity with the acceleration times the time accelerating to find the final velocity. But in the end it was wrong. What did i do wrong?
 
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huynhtn2 said:
The slope is inclined at 48.4°
Do you mean that the slope is declined at 48.4^\circ[/tex] (or that the slope falls at an angle of 48.4^\circ[/tex] below horizontal)?<br /> <br /> Or, is the skier truly jumping onto an uphill slope?
 
Heres a picture of it:
 

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zgozvrm said:
Do you mean that the slope is declined at 48.4^\circ[/tex] (or that the slope falls at an angle of 48.4^\circ[/tex] below horizontal)?<br /> <br /> Or, is the skier truly jumping onto an uphill slope?
<br /> <br /> Here is a picture of it:
 

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