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Finding maximum Height

  1. May 25, 2016 #1
    1. The problem statement, all variables and given/known data
    A soccer ball is kicked at a 45° angle. If the ball is in the air for 3 s, what is the maximum height achieved

    2. Relevant equations
    y=viy + 0.5 x ay x t

    3. The attempt at a solution
    i used trig to get the vertical and horizontal components i got 2.2 for each side (as there the same with a 45 degree angle)

    i used y=0 + 0.5 x (-9.81) x (3s)

    is this right ^^
  2. jcsd
  3. May 25, 2016 #2
    your equation is not correct.
  4. May 25, 2016 #3
    i used y=0 + 0.5 x (-9.81) x (1.5s)Square root

    is what i meant to put
  5. May 25, 2016 #4
    I think you mean squared, not square root.
  6. May 25, 2016 #5
    yes, sorry
  7. May 26, 2016 #6


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    You got 2.2 what? You don't know the velocity, so how can you get the velocity components?
    What does your y0 represent?
  8. May 26, 2016 #7
    y=o represents initial velocity. but it is not known like you said, so that is how i was able to get 11.05 but setting initial to 0
  9. May 26, 2016 #8
    The initial velocity cannot be 0. Otherwise, with gravity being the only accelerator the projectile would go down.
  10. May 26, 2016 #9
    It's possible to calculate your initial velocity though, assuming the projectile lands on level ground. Can you modify your relationship above to help you find this?
  11. May 26, 2016 #10


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    Setting it zero will get you nowhere. Create an unknown for it, v.
    Your equation for y is suitable for a vertical displacement, not a vertical velocity. In terms of v, what would the vertical displacement be at time t? What is the vertical displacement when it lands?
  12. May 26, 2016 #11
    I am trying to find displacement though
  13. May 26, 2016 #12


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    You are asked to find, eventually, the vertical displacement from launch to the highest point. I asked you what the vertical displacement is from launch to where it lands.
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