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Qualitative Explanation of Range

  1. Apr 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Explain qualitatively how it is possible for the projectiles to have the same range despite having different launch angles


    2. Relevant equations
    x = voxt


    3. The attempt at a solution
    A projectile's range is determined by its x component of the initial velocity and the airtime. A projectile can either have a large x-component of its initial velocity, and a short airtime, or a small x-component of its initial velocity and a long airtime. Therefore, they will have the same range.

    Is there a better explanation than this?
     
  2. jcsd
  3. Apr 7, 2014 #2
    Qualitatively:

    You have to imagine the fact that in let's say a 2-dimensional plane where the projectile flies, there are two perpendicular directions. Because of this fact, the y-component of velocity is completely independent of the x-component. It only controls motion in the vertical direction; try thinking of each component as if the other were set to zero.

    For example, if [itex]v_{ox}[/itex] were set to 0, the projectile would go up, then fall right back down. Now if you slowly increase [itex]v_{ox}[/itex] you will basically be spreading out the rise and fall over more space. The rise and fall will still take exactly the same time as if there were no x-component. In the case where there is an x-component it just will not land where it started!


    Mathematically:

    The same initial x-component of velocity will lead to the same range, because they correspond to different angles. For example let's say you have two different projectiles with the same initial x-velocity:

    [itex]
    v_1=[\hat{x}+\hat{y}] m/s
    [/itex]

    [itex]
    v_2=[\hat{x}+\sqrt{3}\hat{y}] m/s
    [/itex]

    If you haven't been introduced to unit vectors, all these [itex]\hat{x},\hat{y}[/itex] symbols mean basically are x-component and y-component respectively.

    So for example with [itex]v_1[/itex] projectile the initial velocity is [itex]1 m/s[/itex] in the x-direction, and [itex]1 m/s[/itex] in the y-direction.

    The [itex]v_2[/itex] projectile has an initial velocity of [itex]1 m/s[/itex] in the x-direction and [itex]\sqrt{3} m/s[/itex] in the y-direction.

    So the second projectile has much more speed in the y-direction than the first projectile. The angle of both can be found from simple trigonometry:

    [itex]
    \theta_1 = \arctan{1} = \frac{\pi}{4} = 45^\circ
    [/itex]

    [itex]
    \theta_2 = \arctan{\sqrt{3}} = \frac{\pi}{3} = 60^\circ
    [/itex]

    So visually you should picture the parabolic path of the projectile, and imagine the two roots of the parabola being it's initial position on the ground, and it's final position. The [itex]45^\circ[/itex] velocity will be much lower to the ground during its flight but will have the same roots as the parabola that had an initial angle [itex]60^\circ[/itex]. All that changed was how high it went, which was determined by the differing y-component. The x-component just changed how much over it went, which was in this case exactly the same!

    I hope this was clear!
     
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