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Kinematics calculation

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data
    a student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 7.6 m/s. The cliff is 38 m above a flat, horizontal beach. Find how long it takes to hit the beach, the velocity at impact, and at what angle below the horizontal the stone is when it lands.


    2. Relevant equations
    y=Vo*sin(initial angle)*t - .5gt^2
    Vox= 7.6 m/s y-yo=38 and g=-9.8m/s^2

    3. The attempt at a solution
    I'm having a hard time getting started, since I only know Vx, the height, and g. I believe the initial angle would be 0 degrees since the student is throwing the stone horizontally and not vertically.
     
  2. jcsd
  3. Sep 13, 2009 #2
    Well, since the problem states that he is throwing the stone horizontally with no reference to vertical motion, the time it takes to hit the ground will be the same as a rock that has been dropped from that height. I myself have only started taking Physics I now, so I am unsure about the other parts of the problem.
     
  4. Sep 13, 2009 #3
    To work out at what time it hits the ground you can just use v^2 = u^2 + 2as, to get it's vertical velocity when it hits the ground, and then use v = u +at to get the time. This is because it's vertical velocity is unaffected by it's horizontal one.

    Likewise it's horizontal velocity remains unchanged throughtout the flight as there is no air resistance. The velocity when it hit's the ground is found by combining the vertical and horizontal velocities it has when it hits the ground. The velocity components form a triangle which you can use to get the angle.

    I would try drawing it out, will help you.
     
    Last edited: Sep 13, 2009
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