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Calculus based kinematics (two dimensions) question

  1. Feb 3, 2015 #1

    RJLiberator

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    Gold Member

    1. The problem statement, all variables and given/known data
    A soccer ball is kicked from the ground at an angle of 30◦ with speed 30m/s. Determine the
    duration of the flight, range and maximal height of the trajectory.

    2. Relevant equations


    3. The attempt at a solution
    1. Break down the velocity vector into x and y components:
    30cos(30) = 15sqrt(3) m/s
    30sin(30) = 15m/s

    r(t) = (0,0)+t(15sqrt(3), 15)+(t^2)/2(0,-9.80)

    This equation represents the position function.

    If I take the derivative to get the velocity function, it becomes
    r'(t) = (15sqrt(3), 15-9.8t)

    When velocity in the y direction is 0, then the ball is at max height
    so 15-9.8t=0
    t=15/9.8 for max height.

    When position of y = 0 the ball reaches its distance and final duration so:
    15t-4.9t^2=0
    t=0 or t= 15/4.9
    clearly it is not t=0.

    since we have time of duration:
    t*velocity in x = range
    15/4.9*15sqrt(3) = 79.5329 m

    max height is found by dividing the time of max range by 2 so thew time of max height = 15/9.8 and then using that as t in the r(t) position function and finding the y-component.
    The y-component comes out to be 11.48m. Seems reasonable.
    15*15/98-9.8/2*(15/9.8)^2=11.48m

    Am I doing this right? My problem is that our calculus class requires no calculators. To me, it seems like this problem is begging the use of calculators unless I just don't simplify things (which would be fine to ). It just seems like I am missing something that'd make the calculations easier.

    Thanks.
     
    Last edited: Feb 3, 2015
  2. jcsd
  3. Feb 3, 2015 #2
    Everything looks right but you really calculated the max height time twice. Once by setting the velocity equation r'y = 0, you solved and got t = 15/9.8

    The other by solving for the full time t = 15/4.9 then dividing by 2.

    These are the same times so you only had to go wih one method. If you want simpler calculations, leave the 9.8 as g so you only have to multiply it through at the way end (or leave it in terms of g if your teacher doesn't mind)
     
  4. Feb 3, 2015 #3

    RJLiberator

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    Gold Member

    Damn, so since everything is correct, I have to assume they either want it simplified via use of a calculator (forbidden generally in this course) or unsimplified and a very whacky answer.

    I guess it is good that I understand the problem :P.

    I will have to talk to my instructor about this.
     
  5. Feb 3, 2015 #4
    You could go a bit further in terms of simplifcation w/o calculator:
    Using y = vot + (1/2)at2
    [itex]y = 15(\frac{15}{9.8}) - \frac{9.8}{2}(\frac{15}{9.8})^2[/itex]

    Cancel out the 9.8's on the right term, collect the squared terms:
    [itex]y = (\frac{15^2}{9.8}) - \frac{1}{2}(\frac{15^2}{9.8})[/itex]

    Factor our (15^2)/9.8 which gives us:
    [itex]y = \frac{225}{9.8}(1 - \frac{1}{2}) [/itex]
    [itex]y = \frac{225}{19.6} [/itex]

    And then a little long division to finish it off? :D
     
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