What is the launch speed required for a skier to reach a height of 11.7 m

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SUMMARY

The launch speed required for a skier to reach a height of 11.7 meters at a 63° launch angle is determined using kinematic equations. The vertical component of the skier's velocity must be calculated first, followed by the total velocity using trigonometric functions. The discussion highlights the importance of understanding vertical and horizontal components of motion in solving projectile problems, emphasizing that these components operate independently.

PREREQUISITES
  • Kinematic equations for projectile motion
  • Basic trigonometry, particularly sine and cosine functions
  • Understanding of vertical and horizontal components of velocity
  • Knowledge of gravitational acceleration (9.81 m/s²)
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  • Study kinematic equations for vertical motion in physics
  • Learn how to decompose velocity into horizontal and vertical components
  • Explore projectile motion problems involving different launch angles
  • Practice calculating launch speeds for various heights and angles
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The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply upward at the end. The sharp upward slope launches them into the air, where they perform acrobatic maneuvers. In the women's competition, the end of a typical launch ramp is directed 63° above the horizontal. With this launch angle, a skier attains a height of 11.7 m above the end of the ramp. What is the skier's launch speed?

i don't know where to begin with this since so little info is given,
 
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This is mostly a 1-D problem.
 
ok, i have the 3 sides of the traingle, but i do now know what to do with them, 13.13, 11.7, 5.96
 
Here's a slightly easier problem you could solve first:
If I throw a ball straight up, how fast do I need to throw it so that it goes up 11.7 meters?
 
EDIT: u are right it is too high.. 15.14m/s is the right one i think
 
Last edited:
That number seems a bit high ...

Now, let's say that I throw the ball at a 45 degree angle with horizontal instead of straight up - how fast do I have to throw it?
 
i have no idea
 
can anyone PLEASE help, i have the answer(16.99), i just don't know how to get the answer
 
What's the vertical component of the velocity of the ball when I throw it?
 
  • #10
10.7m/s ??
 
  • #11
Joules23 said:
10.7m/s ??
It would only go up about 5m if that was the vertical component.
 
  • #12
Vy of the ball at 45degrees or Vy of the ball thrown straight up? wow if every problem is going to take me this long i think ill be done next month
can you give me a formula or something for the actual problem
 
Last edited:
  • #13
anyone? I am completely lost
 
  • #14
Joules23 said:
Vy of the ball at 45degrees or Vy of the ball thrown straight up? wow if every problem is going to take me this long i think ill be done next month
can you give me a formula or something for the actual problem

The way I would do this problem is:
1. Figure out the vertical component of the skiers velocity at the end of the ramp.
2. Use trig to determine the total velocity.
 
  • #15
How would i find the vertical component of the skiers velocity, when it does not give the velocity

IS 11.7 the vertical component?
 
Last edited:
  • #16
The horizontal and vertical components are independent.
 

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