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Finding Launch Velocity given time

  1. Oct 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi,
    So the question asks me to find the launch velocity of a bullet that was shot up at a 90 degree angle. The bullet then came back down at the same height that it was shot at in 1.37 seconds.

    I'm not sure if I'm using the right formula, or even plugging the right numbers in.
    2. Relevant equations

    3. The attempt at a solution
    I used the formula dy= vy.t + ay.t /2
    dy = 0
    vy = launch velocity
    t = 0.685 (because i divided it by 2 for the max?)
    and a = -9.8 m/s/s

    Or would I have to use the formula a=(vf - vi)/2 ? with vf = 0 ?
    Sorry, I'm really confused.

    Thanks! :)
     
  2. jcsd
  3. Oct 20, 2012 #2

    PeterO

    User Avatar
    Homework Helper

    Since you took dy = 0, you were considering the tip from start, back to ground level, so you would use the whole time.
    NOTE: 1.37 seconds does not mean the "bullet" went very high? Is this actually a child's toy type of gun or is that time a bit short. A bullet from a real gun is likely to take 30 - 60 seconds to come back down.
     
  4. Nov 2, 2012 #3
    It was a fake bullet. But thank you so much! :)
     
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