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Height of a projectile at a certain point

  1. May 21, 2007 #1
    1. The problem statement, all variables and given/known data
    A soccer player kicks a soccer ball at a speed of 33.5 m/s at an angle of 10.6 degrees at the goal 25.0m away.

    2. Relevant equations

    What is the height of the ball when it reaches the goal?

    3. The attempt at a solution

    Was able to calculate the final velocity but need a little help setting up this one
  2. jcsd
  3. May 21, 2007 #2


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    Homework Helper

    Come on, please show your work.
  4. May 21, 2007 #3
    [tex]d = 1/2(v_o+v_f)t[/tex]
    [tex]d = v_{o}t+1/2at^2[/tex]
    [tex]v_f = v_o+at[/tex]
    [tex]v_{f}^2 = v_{o}^2+2ad[/tex]

    Those equations are a good place to start^^

    Remember there's no acceleration in the "horizontal" direction
    Last edited: May 21, 2007
  5. May 21, 2007 #4
    Work so far

    Ok... so far I calculated

    sr = (2(33.5 m/s)^2)/(9.81 m/s^2) cos10.6 sin10.6

    sr = 41.4m

    this will give me the range if the ball struck the goal at the same height it was kicked from. I cant figure out how to calculate its height 25.0m away.
  6. May 21, 2007 #5
    I don't see how finding the total range of the ball will help you...

    [tex]d = v_{o}t+1/2at^2[/tex]

    If you tried to solve this for the horizontal direction then [tex]a=0 m/s^2[/tex]. So you're left with [tex]d = v_{o}cos(10.6)t[/tex]. You have a displacement (the goal) and you know the intial velocity so you can figure out how long it will take to reach the goal...
    Last edited: May 21, 2007
  7. May 21, 2007 #6
    I calculated the time to be 1.71s, when plugged into the equation to solve for "d" it gives me a height of 10.35m. Looks like a feasible answer (hope I did it right lol). Thanks so much for helping.
  8. May 21, 2007 #7


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    10.35 m?!? That's pretty high! How did you arrive at that value for t?
  9. May 21, 2007 #8
    Vf = Vo + at

    22.99m/s = 6.16m/s + 9.81m/s^2 ( t )

    t = 1.71s
  10. May 21, 2007 #9
    You don't know what the velocity will be when the ball reaches the goal unless you previously know the time. Where did you get 22.99 m/s from?
    Last edited: May 21, 2007
  11. May 21, 2007 #10
    [tex]v_{f}^2 = v_{o}^2+2ad[/tex]

    Vf^2 = (6.16m/s)^2 + 2(9.81m/s^2)(25m)

    Vf = 22.99m/s

    wrong way I guess
  12. May 21, 2007 #11
    The key to projectile questions are knowing which formula to use, and 9 times out of ten, you need to work out the horizontal components before you can start on the veritical (due to the effect of acceleration on vertical velocity over time).

    As we have estbalished however, the is a link between the two .. time.

    List out what you know about the hoiztonal components, and choose a formula that will be easiest to find time with.

    Then you can start working out the vertical elements.
  13. May 21, 2007 #12


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    The key is that the horizontal component of motion is at constant velocity, and that the vertical component is at constant acceleration (due to gravity). So separate your two components. Using [tex] x = v_{x} t [/tex] solve for t and then use that in the equation for uniform acceleration.
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