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Projectile motion when you are only given time and range

  1. Mar 13, 2017 #1
    1. The problem statement, all variables and given/known data
    I am very lost and would love help. Given in the description is;"A golf ball leaves a golf club at an angel 30 degrees above the horizontal and lands on the green 150 meters away initial position." With this information i am supposed to find the balls initial velocity, how long it was in the air and how high it rises

    3. The attempt at a solution
    to find the initial velocity i did delta r = [vi^2 * Sin (60) ]/ 9.81 = 41.22.
    not 100% sure this is correct but its what i have so far and i have no idea where to go from here
     
  2. jcsd
  3. Mar 13, 2017 #2

    cnh1995

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    Your calculation of initial velocity is correct.

    You need to find the time of flight (T) and maximum height (H). Do your notes contain the formulas for T and H ?
     
  4. Mar 13, 2017 #3
    okay thanks and they don't contain either of those formulas and i don't see either in my notes as well.
     
  5. Mar 13, 2017 #4

    cnh1995

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    Well, that's odd!

    Anyway, you can look them up or you can derive them on your own if you applied your knowledge of vectors and kinematic equations.
    What are the x and y components of initial velocity? What is the acceleration in x and y directions?
     
  6. Mar 13, 2017 #5
    for the x and y components of vi would it be 41.22cos30 and 41.22sin30?
     
  7. Mar 13, 2017 #6

    cnh1995

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    Yes.
    What about acceleration?
     
  8. Mar 13, 2017 #7
    no acceleration in the x and a acceleration of -9.81 in the y?
     
  9. Mar 13, 2017 #8

    cnh1995

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    Correct.
    Can you write the relevant kinematic equations for finding T and H?
     
  10. Mar 13, 2017 #9
    im sorry but i have no clue and im not able to find it anywhere.
     
  11. Mar 13, 2017 #10

    cnh1995

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    What is the horizontal dispalcement? Which component of velocity is responsible for it?
     
  12. Mar 13, 2017 #11
    vix and time dictate horizontal displacement
     
  13. Mar 13, 2017 #12

    cnh1995

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    Yes.
    So you can find time from this. You know Vx and horizontal displacement.
     
  14. Mar 13, 2017 #13
    So if the horizontal displacement is 150 and the vx is 35.7 it would be 150/35.7=time=4.2 seconds?
     
  15. Mar 13, 2017 #14

    cnh1995

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    Yes.
     
  16. Mar 13, 2017 #15
    and the equation for height, would that be vy + (0.5*9.81*4.2^2) ?
     
  17. Mar 13, 2017 #16

    cnh1995

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    Careful there..
    At what time will the ball be at the maximum height?
     
  18. Mar 13, 2017 #17
    at half the time so 2.1 seconds?
     
  19. Mar 13, 2017 #18

    cnh1995

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    Right.
     
  20. Mar 13, 2017 #19
    im confused now, how would that change the equation
     
  21. Mar 13, 2017 #20

    cnh1995

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    You have taken t=4.2s..
     
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