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Jumping a Crevasse (Projectile Motion)

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data
    A mountain climber jumps a crevasse by leaping horizontally with speed vo. If the climber's direction of motion on landing is [tex]\vartheta[/tex] below the horizontal, what is the height difference between the two sides of the crevasse?


    2. Relevant equations
    I'm not sure, but I think it's y=h-1/2gt2


    3. The attempt at a solution
    My first thought was to just manipulate the equation to give h=y+1/2gt2, but I'm sure that's not right. What's really confusing me is [tex]\vartheta[/tex], because I'm not sure how knowing that is pertinent.
     
  2. jcsd
  3. Jan 24, 2010 #2

    rl.bhat

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    Hi Annan. welcome to PF.
    In the projectile motion horizontal velocity vo remains constant. But the vertical velocity increases. And it is given by v = gt......(1)
    If θ is the angle of landing then tanθ = v/vo.....(2)
    Difference in height Δh = 1/2*g^t^2 ........(3)
    Using eq. 1 and 2, eliminate t and v and find Δh in terms of vo, g and tanθ.
     
  4. Jan 25, 2010 #3
    This is probably a dumb question, but in y=h-1/2gt^2, is y the final height, and therefore equal to 0?
     
  5. Jan 25, 2010 #4

    rl.bhat

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    Not necessarily. In the problem you have to find (h - y) which is Δh = 1/2*g*t^2
     
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