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Horizontal Projectile Motion Problem

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A mountain climber jumps a crevasse by leaping horizontally with speed [tex]v_{o}[/tex]. If the climber's direction of motion on landing is [tex]\Theta[/tex] below the horizontal, what is the height difference [tex]h[/tex] between the two sides of the crevasse?


    2. Relevant equations
    [tex]x=v_{o}t[/tex]
    [tex]v_{x}=v_{o}[/tex]
    [tex]y = h - \frac{1}{2}gt^{2}[/tex]
    [tex]v_{y}= -gt[/tex]
    [tex]v_{y}^{2} = -2g\Delta y[/tex]
    [tex]y = h - (\frac{g}{2v_{0}^{2}})x^{2}[/tex]
    [tex]x = v_{o}\sqrt{\frac{2h}{g}}[/tex]

    3. The attempt at a solution
    I started off by drawing the following diagram:
    http://img191.imageshack.us/img191/2616/physicss.jpg [Broken]

    Afterwards, I used the formula [tex]y = h - \frac{1}{2}gt^{2}[/tex] to solve for [tex]h[/tex], and I ended up with [tex]h=\frac{1}{2}gt^{2}[/tex]. I then used the formulas [tex]x=v_{o}t[/tex] and [tex]x = v_{o}\sqrt{\frac{2h}{g}}[/tex] to come up with the formula [tex]t = x\sqrt{\frac{2h}{g}}[/tex]. I then substituted this formula into [tex]y = h - \frac{1}{2}gt^{2}[/tex], made [tex]y=0[/tex] and simplified the equation until I got [tex]x=1[/tex]. So, now I know what the [tex]x[/tex] value is when [tex]y=0[/tex]. The problem is I'm stuck right here and don't know how to find the height. Can someone point me in the right direction please?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 25, 2010 #2
    You cant get a particular solution since you arent given any values. So h = 0.5gt^2 is correct.

    Remember when dealing with these types of questions to work out the vertical and horizontal probelms seperately. The climber could have jumped 1m or 500m horizontally, but it wouldnt have changed how long he would have been in the air for nor the height of the crevasse.
     
  4. Jan 25, 2010 #3
    So I shouldn't specify an x value in my answer?
     
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