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Projectile motion and initial speed problem

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data

    A projectile returns to its original height after 4.08 seconds, during which time it travels 76.2 meters horizontally. If air resistance can be neglected, what was the projectile's initial speed?
    (Use g = 9.80 .)


    2. Relevant equations



    3. The attempt at a solution

    Hey guys I already solved the problem I just want to get some feedback on whether my breakdown of what I did was correct.

    So what I did first was split the problem into its X and Y ( horizontal and vertical components).

    76.2m=v0 (4.08s)
    vx=18.67m/s

    Then since the projectile returned to its original height this means that its displacement or change in position will be zero. So I can solve for its initial velocity , by doing the following

    0=-.5(9.8)(4.08)^2+v0(4.08)
    v0=20m/s

    Here is where I get confused I know I have to do the magnitude formula to get the intial speed of 27.4. squaroot ((18.67)^2+(20)^2))

    My question is why do I have to do that?
     
  2. jcsd
  3. Feb 4, 2012 #2

    BruceW

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    Homework Helper

    the velocity is a vector and the speed is the magnitude of the velocity. So the reason you have to do that magnitude formula, is just because that is mathematically the way to get the magnitude.

    For an intuitive explanation, you can think of the velocity as a line in 2d space, then the speed is the length of this line, so you use Pythagoras' theorem to find the length in terms of two perpendicular lines (which are analogous to the components of the velocity). Of course, the velocity vector isn't actually a line in 2d space, but this gives a way of imagining how it works, intuitively.
     
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