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Projectile motion, deriving equation

  1. Jan 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Derive equations (5) and (6). Calculate at what angles the time and range are maximized.

    (5) t= ((2 * Vo) / g) * sin[tex]\theta[/tex]
    (6) Range = x1-xo = ((2 * Vo)/g)(sin[tex]\theta[/tex])(cos[tex]\theta[/tex])= (Vo^2/g)(sin2[tex]\theta[/tex])

    sorry about the confusing format. i'm new to latex and new to this forum

    2. Relevant equations

    (1) Vox = Vo * cos[tex]\theta[/tex]
    (2) Voy = Vo * sin[tex]\theta[/tex]
    (3) Range = X - Xo = Vox * t
    (4) Y - Yo = Voy * t - (1/2)gt^2

    3. The attempt at a solution

    I have no clue how to derive an equation. This is my first physics class i've taken ever and it was never taught by my professor on how to derive an equation.

    i hope you guys can bear with the tiny amount of knowledge i have. i tried googling instructions and ways on how to derive an equation but i still don't understand the concept. thanks!
  2. jcsd
  3. Jan 20, 2010 #2
    Well to derive (5) I suggest you use equation (4). The change in Y, which you have written as Y-Yo would be zero if we assume a projectile lands at the same Vertical height from which it is launched. Then if you move the first term in the (4) (has t in it) over to the other side, you take it from there. You are really just substituting and doing some algebra.

    You do not have all the kinematic equations in your list that are available and might help in the last one (6)
  4. Jan 24, 2010 #3
    thanks! i think i got it the whole derive thing down now.

    for the 2nd part where it asks, "Calculate at what angles the time and range are maximized.", am i just finding theta for both equations (5) and (6)?
  5. Jan 24, 2010 #4

    So for spending the longest time in the air it should be fairly obvious all your velocity should be devoted to the vertical direction. So you really know the angle before you derive it. This should give you a "target" as help in checking if you derived it properly.

    The range is less obvious as you need to have as much horizontal velocity as you can get for the longest amount of time. A fairly large t and a fairly large Vx combined. So this is really a playoff between cos and sine. One way to think about this is you want an angle that if you multiplied sin of theta x cos of theta (theta being some angle) gives you the largest number. Now this is not a derivation of the angle. But again it gives you a somewhat practical way of thinking about what the answer should be.

    And you can keep going with this stuff. Dervive the formula for the angle in which you would go as high (vertical max) as you would go far (horizontal or range). Sometimes it is good a exersize in playing these little games. But sometimes students just need to get the problem finished and move on.
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