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Projectile motion of a pebble

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data

    a tourist is climbing high up the pyramid of cheops, which has sloping faces that make and angle of theta with the ground. The tourist throws a pebble with initial speed v is a direction perpendicular to one of the faces. the the height at which the peble hits the pyramid below the tourist

    2. Relevant equations



    3. The attempt at a solution

    y = vo + vyt + .5at
    y = vtsin(90 - theta) + 4.9t


    what else can i plug in. im lost
     
  2. jcsd
  3. Sep 3, 2009 #2

    rl.bhat

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    Homework Helper



    The above equations are wrong. Check them.
     
  4. Sep 3, 2009 #3

    berkeman

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    Staff: Mentor

    Draw a sketch of the problem. Think both in the x and y directions as you write your equations. The pebble will have a constant velocity in the x direction of ____? Given that info, you can calculate how far down the face the pebble will hit as a function of time.

    Then think in the y direction, and write an equation that describes the height y above the launch point the pebble will be as a function of time. Equate the two y's, and that's where and when the pebble will hit the face...
     
  5. Sep 9, 2009 #4
    vx = vcos(90-[tex]\theta[/tex])
    vy = vsin(90-[tex]\theta[/tex])

    x = vocos(90-[tex]\theta[/tex])t

    is this usefull tan[tex]\theta[/tex] = y/(vcos(90-[tex]\theta[/tex])t)
     
  6. Sep 10, 2009 #5
    it seems i have 4 variables, x,v,t[tex]\theta[/tex] can i get a clue on how to eliminate them
     
  7. Sep 10, 2009 #6
    ok made some progress but still a little confused

    x = vtcos(90-[tex]\theta[/tex]) therefore t = x/(vcos(90-[tex]\theta[/tex]))

    y = vsin(90-[tex]\theta[/tex]))(x/(vcos(90-[tex]\theta[/tex]))) - 4.8(x/(vcos(90-[tex]\theta[/tex])))

    y = xtan(90-[tex]\theta[/tex])) - 4.8x/(vcos(90-[tex]\theta[/tex]))) projectile

    now i found the y based on the angle

    tan[tex]\theta[/tex] = -y/x, therefore y = -xtan[tex]\theta[/tex] so i set both y's equal

    -xtan[tex]\theta[/tex] = xtan(90-[tex]\theta[/tex])) - 4.8x/(vsin[tex]\theta[/tex])

    is this correct so far
     
    Last edited: Sep 10, 2009
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