1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

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


    User Avatar

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook