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Kinematics of a particle

  1. Oct 29, 2016 #1

    Kajan thana

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    Gold Member

    1. The problem statement, all variables and given/known data
    A ball is thrown horizontally wit speed 20m/s, from the top of the building which is 30m high.
    Find the time the ball takes to reach the ground?

    2. Relevant equations
    S= ut+0.5at^2
    3. The attempt at a solution
    S=-30
    u=0
    A=9.8
    T=?

    When I pluck the values I will get t^s=-30/4.9 but when we square root we will get complex number, but if I take the displacement as +30 then it will give me a positive value, so I am confused on why the s is not negative.

    If I say the displacement where the ball is on top of the building, then the displacement is 0 but if it goes down then it should be negative value.
     
  2. jcsd
  3. Oct 29, 2016 #2
    I think if I was working this problem, to make it easiest, I would define the down direction as positive (that way acceleration and displacement will both be positive) and the top of the building as x=0. Or you could define down as negative and acceleration and displacement would be negative. But it's for sure that if you throw a ball off of a building, the time required for the ball to hit the ground will be a positive value.
     
  4. Oct 29, 2016 #3

    Kajan thana

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    Gold Member

    Perfect, I thought about this, but I did not know if this will be right.
    Thanks.
     
    Last edited: Oct 29, 2016
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