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Distance traveled by a Projectile?

  1. May 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Horizontal distance traveled by a projectile fired from the ground can be modeled by d= (v^2sin2x)/g, where v is initial velocity, x is the angle in degrees, and g is acceleration due to gravity. (This model neglects air resistance and assumes a flat surface)

    a) A baseball is hit at an initial height of 5 feet with initial velocuty of 100 ft/s at 40 degrees to the ground. Will the ball clear a 5 ft fence 310 ft. from home plate?
    b) What angle will maximize the horizontal distance traveled and why?
    c) Will the baseball clear the fence if hit at the angle from part (b) above?


    2. Relevant equations

    (v^2sin2x)/g

    3. The attempt at a solution
    a) I dont get how it gives you an initial height of a fence, when height isn;t even mentioned in the given equation?
    b) (v^2sin2x)/g = 310
    sin2x = 310 *9.81 / 100^2
    sinx = 540
    x = 21.4 degrees
    c) Yes?
     
  2. jcsd
  3. May 5, 2008 #2
    a) sorry I'm not familiar with baseball (and english isnt my native language), what do you mean by "clearing a fence" ?
    I assume you either mean that the baseball passes above, or hits the fence. Either way :
    The baseball is hit at the same height than the fence. The equation d= (v^2sin2x)/g applies to projectile fired from the ground, so it says in your question, it means that it gives the distance the projectile has reached when hitting the ground, so the same level. So, in fact, the height of launch doesnt matter, you can still use this equation, but it won't measure the distance til fall on the ground but til reach of the launch height once again.

    Here we launch our baseball from a 5 ft height, and we need to know if it will either pass above or hit a 5 ft fence. It means that if the distance found by the equation, the distance til next time the ball has an altitude of 5 ft, is superior to the distance of the fence, the ball will pass above. But if the distance found is inferior to 310 ft, we have a problem because either the ball will hit the fence, or hit the ground too soon. Maybe it can bounce on the ground, roll, and so on. So I would say that it would be more logical for such an exercise to ask if the ball will pass above. Note : this is just about my understanding, in fact you all understand the question lol, I'm talking to myself !

    b) The question is about which angle allows the maximum distance. This has nothing to do with the fence ! Look, you have an sinus in your equation, and you know that the sinus has values between 0 and 1 (for angles between 0 and 180°). What value is needed for a maximum distance ? which launch angle allows such a value ?

    c) once you have your angle, you can get the corresponding distance, just compare it to the distance to the fence
     
  4. May 5, 2008 #3
    A) You don't need (v^2sin2x)/g to solve part a). Break the displacement vectors into components.

    B) sin(x) is maxed when it is equal to 1 which happens when the angle is 90 degrees (pi/2 rad). However you have sin(2x)

    C) You cannot do this without parts a) and b).
     
  5. May 5, 2008 #4

    djeitnstine

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    Gold Member

    This question simply asks will the ball travel 310 ft or more horizontally. So you only need to get its horizontal component

    b) x = 22.5 degrees , redo your calculation

    Also I do not understand your equation, is that v[tex]^{2sin(2x)}[/tex] or v*(2sin(2x)) because I saw you divide by 100*2 but you seem to have it as 100[tex]^{2}[/tex]

    If it is indeed multiplied then answer should be no to c
     
  6. May 5, 2008 #5

    I'm pretty sure it's not 22.5 degrees either.
     
  7. May 5, 2008 #6

    djeitnstine

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    Gold Member

    oops you are correct Sine is maxed when x = pi/2 or 90 degrees. Since he has 2x = pi/2 then it should be x = pi/4 which is 45 degrees.
     
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