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Shooting Rifles from Platforms - Projectile question

  1. Sep 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi all! What a great site, thanks in advance for any help! My question is written:
    The six figures below show rifles that are being fired horizontally, i.e. straight out, off platforms. The bullets fired from the rifles are all identical, but the rifles propel the bullets at different speeds. The specific speed of each bullet and the height of the platform are given. All of the bullets miss the targets and hit the ground.

    Rank these bullets, from longest to shortest, on the basis of how long it takes a bullet to hit the ground. That is, put first the bullet that will take the longest time from being fired to hitting the ground, and put last the bullet that will take the shortest time.

    Following this are six pictures of identical trees, each with a platform and a little man with a rifle. There are distance measurements beside the trees for the height of each platform, and speeds above each rifle.

    The first tree has a platform height of 20 m, and a bullet speed of 1200 m/s.
    The second tree has a platform height of 40 m, and a bullet speed of 500 m/s.
    The third tree has a platform height of 40 m, and a bullet speed of 800 m/s.
    And so on.

    2. Relevant equations
    My instinct is to use the equation:
    [itex]t=\sqrt{\frac{2d}{a}}[/itex]

    I know the distance and the acceleration (-9.8 m/s^2), so i get an answer, the problem is if I use this equation to get the time, I'm not using the bullet speeds I'm given. Is that correct?


    3. The attempt at a solution
    [itex]t= \sqrt{\frac{2(-20 m)}{-9.8 m/s^2}}[/itex]
    t=2.0s

    Is this right? Thanks again for your help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 8, 2013 #2
    You can use this equation to relate time and vertical displacement if the initial vertical velocity is zero. Is it?
     
  4. Sep 8, 2013 #3
    Yes, I think is. The horizontal velocity is the speed of the bullet fired from the rifle. That makes the vertical velocity zero. (I think.)
     
  5. Sep 8, 2013 #4
    Then you should not be surprised that speed does not affect your results.

    Would anything change if the rifles were at an angle with the horizontal?
     
  6. Sep 8, 2013 #5
    Yes, then I would have to use the velocity for the calculation, not the distance. Hmmm, good point!
     
  7. Sep 8, 2013 #6
    You would have to use both. Anyway, I think you know how to solve this problem as stated.
     
  8. Sep 8, 2013 #7
    Thank you so much, you're right, I think I do know what to do now!
     
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