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What is the point of angling projectiles?

  1. Dec 25, 2012 #1
    1. The problem statement, all variables and given/known data
    If the horizontal velocity remains the same, whether angled or not, I dont quite understand why we bother launching projectiles upwards on an angle?

    Would it be better just to launch them horizontally ?


    2. Relevant equations

    d= 1/2gt^2....I think?


    3. The attempt at a solution

    I know there must be a reason but it doesn't seem any extra distance is being achieved if the the projectiles horizontal distance is the same regardless of it is angled or not..

    thinking of the Olympics for example, I cant imagine them throwing the javelin horizontally , but I don't quite get the idea behind making the launching on an angle
     
  2. jcsd
  3. Dec 25, 2012 #2

    gneill

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    Projectiles don't fly so well through the ground.

    Unless you can find a way to turn off gravity, any projectile you launch (that doesn't leave the planet entirely) will eventually hit the ground. Usually that brings a sudden halt to the proceedings. The questions to ask yourself are, will it hit the ground sooner or later if I launch it at an angle? and, is there an optimum angle for furthest distance given a fixed launch speed?
     
  4. Dec 25, 2012 #3
    Well, he said in his hypothetical example that the horizontal velocity component is constant, so the optimal angle would be very close to 90 degrees.

    OP, note that it is pretty hard to maintain constant horizontal velocity at different angles. Usually the object starts at an initial velocity and you can find the horizontal and vertical components of the velocity depending on the angle with basic trigonometry. An angle of 90 degrees means that your entire velocity is in the vertical component. An angle of 0 degrees means that your entire velocity is in the horizontal component.
     
  5. Dec 26, 2012 #4

    SteamKing

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    If the optimum angle is 90 degrees, the projectile goes straight up. Remember to duck when it comes back down.

    In order to maximize the horizontal distance traveled, or range for all you ballistic fans, the projectile is launched at an angle to the horizontal. Without discussing drag or wind, the optimum angle of launch for maximum range is 45 degrees to the horizontal.

    To see why an angle of launch of zero degrees is not desirable for maximum range, consider the following. If a projectile is launched at zero degrees to the horizontal, it will travel in the horizontal direction parallel to the ground, but it is also dropping toward earth under the influence of gravity. Since there is no vertical component of velocity, the projectile will reach the ground in the same time it would take an object that was dropped from the same height at which the projectile was originally launched. In most instances, this height is very small, hence the time of flight of the projectile is a few seconds.

    If, on the other hand, a projectile is launched at an angle to the horizontal, it travels upward at a certain velocity and gains altitude during flight. The influence of gravity is still acting to slow the upward travel of the projectile, and once the vertical component of velocity reaches zero, the projectile falls back to earth. One can see that the projectile launched at an angle spends more time aloft than one launched at a zero angle to the horizontal. Consequently, the angled projectile has a greater range.
     
  6. Dec 26, 2012 #5

    CWatters

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    Is that a problem statement or ??

    I ask because the horizontal velocity doesn't normally remain the same if the launch angle changes. It normally changes according to Cos(θ).
     
  7. Dec 26, 2012 #6

    CWatters

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    You can prove that a 45 degree angle is best (ignoring wind or drag)....

    Let...

    θ = launch angle
    u = initial velocity
    x = horizontal distance from launch
    y = vertical distance from launch

    Vertically..
    y = uSin(θ)t - (1/2)gt2 .....................(1)

    Horizontally..
    x = uCos(θ)t .....................................(2)

    What we are interested in is the value of t and then x when the object reaches the ground again (eg when y=0) so solve (1) for t when y=0

    0 = uSin(θ)t - (1/2)gt2
    divide by t
    0 = uSin(θ) - (1/2)gt
    rearrange
    t = 2uSin(θ)/g

    Substitute for t in (2)

    x = uCos(θ) * 2uSin(θ)/g
    or
    x = (u2/g) * 2uCos(θ)Sin(θ)

    Rembering that
    2Sin(θ)Cos(θ) = Sin(2θ)
    gives

    x = (u2/g) * Sin(2θ)

    So we want to know at what angle θ is x a maximium...

    Sin(2θ) is a max when θ = 45 degrees.
     
  8. Dec 26, 2012 #7
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