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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

    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!


    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:

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

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

    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:

    \theta_1 = \arctan{1} = \frac{\pi}{4} = 45^\circ

    \theta_2 = \arctan{\sqrt{3}} = \frac{\pi}{3} = 60^\circ

    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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