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← 06ps-10 Ski Jumping

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Showing Revision 1 created 10/27/2012 by Amara Bot.

  1. Let's start off with this problem--let's pick a location to define

  2. a potential energy U to be equal to 0.
  3. I'm going to choose just this line here, a place where the skier will land
  4. and where all of our heights are measured from just to make things nice and easy.
  5. The first part of the problem, I want to figure out what the speed of the skier is
  6. at this point right here when he finally just leaves ramp
  7. because that is just a projectile motion problem, and we know how to do that from unit 2.
  8. I want to figure out using energy techniques.
  9. What his kinetic energy and therefore what his speed will be when he's leaving the ramp right here.
  10. Now, we know from conservation of energy that the skier's kinetic and potential energy
  11. must add up to the same value in both his starting position here and when he's just leaving the ramp.
  12. We can actually get rid one of these four variables already.
  13. At the beginning of the jump, the skier isn't moving yet, so we know his kinetic energy must be 0.
  14. Doing a little Algebra, I find that his final kinetic energy should be equal to the difference
  15. between his original potential energy and his final potential energy.
  16. And doing a little algebra and plugging in the values we were given,
  17. we find that the skier's velocity is 40 m/s when he's leaving the ramp here.
  18. Now that I know the speed of the skier at the tip of the ramp,
  19. I can now calculate using kinematics how far away from the ramp he will land.
  20. Note that in the previous part of the problem, 40 m/s was what we called
  21. the final speed, and it was for that part.
  22. It was what the skier's speed was at the end of the first part of the problem
  23. when he was just on the tip of the ramp.
  24. But now, that 40 m/s is the original speed for the second part of the problem
  25. when the skier is flying through the air.
  26. That's why I'm using this convention of calling this V0 now.
  27. The first thing I'm going to do is break this original velocity
  28. into x and y components based on this angle α.
  29. And in doing so, I find that the original Vx is 34.64 m/s, and the original Vy is 20 m/s.
  30. Now, I'm going to use this kinematic equation to solve for the time t
  31. when the skier lands on the ground and that's when his change in y would be -10 since 10 appeared
  32. and then he goes down to 0, a change of -10, and doing a little algebra, I write this quadratic equation.
  33. Now, it doesn't look like I can factor this quadratic equation,
  34. so I'm going to have to use the quadratic formula.
  35. And plugging these numbers into the quadratic equation,
  36. I get two solutions, t=-0.45 s and t=4.45 s.
  37. Now, what's the deal with this negative time one?
  38. Well, if we look back at our graph, we can actually imagine
  39. the skier going backwards in time on his trajectory, prodopdopdop.
  40. And he would actually intercept the y equal 0 line right about here.
  41. Our solution is actually still correct. It's not quite what we're looking for in this problem.
  42. Instead we're going to focus on this positive solution, t=4.45 s, and that sounds reasonable.
  43. It sounds like it might take about 4.5 s for the skier to go in the air and land back on the ground.
  44. Okay. Now, I'm going to use this equation to figure out how far in the x direction
  45. the skier travels during that time he is in the air.
  46. Now keep in mind there's no acceleration in the x direction so this term goes to 0,
  47. and we are left with just Δx=V0x*t.
  48. And plugging in the numbers we were given, we got that the final change in x.
  49. The distance d that the skier lands away from the ramp is 154.1 m. If you got this answer, great work