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Football Projectile Problem

  1. Oct 5, 2005 #1
    A football player throws a football with an initial upward velocity component of 18.0 m/s and a horizontal velocity component of 25.0 m/s. (a) How much time is required for the football to reach the highest point of the trajectory? (b) How high is this point? (c) How much time (after being thrown) is required for the football to return to its original level? How does this compare with the time calculated in part (a)? (d) How far as it traveled horizontally during this time?

    (a) Ok so I know that [itex] v_{x}_{0} = 25.0 m/s, v_{y}_{0} = 18.0 m/s [/itex]. So for part (a) we know that the vertical component of velocity [itex] v_{y} = 0 [/itex]. So would I use [itex] t = \frac{v_{0}\sin\theta_{0} - v_{y}}{g} [/itex] where [itex] v_{y}_{0} = v_{0} = 18.0 m/s, v_{y} = 0, \theta = 90, g = 9.8 m/s^{2} [/itex]?

    (b) Would I use [itex] y = (v_{0}\sin\theta_{0})t - \frac{1}{2}gt^{2} [/itex] and plug in the time?
    (c) wouldnt the times for part(a) and part (c) be equal neglecting air resistance?
    (d) I would just find the time for the football to travel the horizontal distance and use the equation [itex] R = (v_{0}\cos\theta_{0})t_{2} [/itex]?

    Thanks
     
    Last edited: Oct 5, 2005
  2. jcsd
  3. Oct 5, 2005 #2
    Remember that v=<25.0, 18.0>m/s. You can either plug that in and find your vertex (similar to what you suggested above) or differentiate and set equal to zero (if you don't know how to take a derivative yet, forget I said that). To find the time it takes for it to hit the ground, you set the y equation equal to zero (do you see why?) You will have two roots to this equation, one being your initial time and the other being the time you need to find. I hope this helps.

    Alex
     
  4. Oct 5, 2005 #3
    yeah I think I got it. Is this correct:

    (a) [itex] t = \frac{v_{0}\sin\theta_{0} - v_{y}}{g} = \frac{18.0 m/s}{9.8 m/s^{2}}, t = 1.1836 s [/itex]

    (b) [itex] y_{max} = (18 m/s)(1.86) - \frac{1}{2}(9.8 m/s^{2})(1.836)^{2} = 16.53 m [/itex]
    (c) [itex] t_{2} = \frac{2v_{0}\sin\theta_{0}}{g} = \frac{ 36 m/s}{9.8 m/s^{2}} = 3.673 s [/itex].
    (d) [itex] R = (v_{0}\cos\theta_{0})t_{2} = (25 m/s)(3.673 s) = 91.83 m [/itex]

    Thanks
     
  5. Oct 5, 2005 #4
    is this correct?
     
  6. Oct 5, 2005 #5

    Fermat

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    Homework Helper

    You upward velocity is 18 m/s, decresaing by 9.8 m/sĀ², so time to reach zero velocity, at which the foorbal is at its maximum height is,

    t = Vy/g
    t = 18/9.8
    t = 1.837 s
    ========

    (b) is correct

    (c) is correct

    (d) is correct
     
    Last edited: Oct 5, 2005
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