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What is the skier's launch speed?

  1. Mar 2, 2004 #1
    Please Help....

    Can Someone please help me with these 2 physics question..i don't know waht equation to use ...can someone please tell me what equation to use..thanks alot..appreciate it.

    1.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 degrees above the horizontal. With this launch angle, a skier attains a height of 12.0 m above the end of the ramp. What is the skier's launch speed?

    2. Suppose the water at the top of Niagara Falls has a horizontal speed of 4.88 m/s just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a 52.3 degrees angle below the horizontal?
  2. jcsd
  3. Mar 2, 2004 #2
    1. [tex]\Delta (v^2) = 2ad[/tex]

    2. As #1, but also requires, [tex]\tan \theta = \frac{v_y}{v_x}[/tex].

    Honestly, I suggest not memorizing a bunch of formulas to do problems like this. It's really easier to just model each problem, because then you do every single problem the same exact way, regardless of what it's asking. All that changes is the value that you read off your model at the end.

  4. Mar 3, 2004 #3
    hey thanks alot..so the number 1 i just use that euqation to find the speed..so i don't need the 63 degree or anything like that..and for number 2..i use the same equation as #1 to find D right...but before i do that i need to find Vy right and that's using the tangent equation right??
  5. Mar 3, 2004 #4
    For #1, close. Remember that some of the initial velocity of the jump is going to be in the horizontal direction, so the vertical velocity will be less than the total velocity. Use a trig identity to separate out the vertical velocity.

    For #2, yes.

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