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Help with homework question

  1. Mar 3, 2009 #1
    1. A tennis ball is thrown from ground level with velocity directed above the horizontal. If it takes the ball 0.50 s to reach the top of its trajectory, what is the magnitude of the initial velocity? You should draw a diagram to show this problem in your explanations.


    i no that Ax=0 Ay=-9.8 i have all the equations just I dont know which to use, once i figure taht out i should be fine any help would be great
     
  2. jcsd
  3. Mar 3, 2009 #2

    LowlyPion

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    Welcome to PF.

    If you know how long it took for acceleration to reduce the velocity to 0 at the top of the trajectory, then you know the vertical component of velocity.

    v = a*t

    You know a and t.

    But without further info about what happens in the x direction or the angle, that's about all you can tell is the vertical component.
     
  4. Mar 3, 2009 #3
    so basically magnitude would be 4.9m/s since your dividing gravity by .5?
     
  5. Mar 3, 2009 #4
    or would it be 9.8 since the total flight path is 1 second... this is where i get confused

    Now for finding the ragne i know the R= (VoCOS(30))T which in this case would be 8.5m

    I know i didnt explain well how i got these figures but its hard to type it all out on a keyboard. any help would be great and thank you in advance
     
    Last edited: Mar 3, 2009
  6. Mar 3, 2009 #5

    LowlyPion

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    This is correct.
     
  7. Mar 3, 2009 #6

    LowlyPion

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    Now that you have revealed that the range is 8.5m then you can use the total time - which is twice the time to max height to determine the Vx. In this case since time is 1 sec, that makes Vx 8.5 m/s.

    So there you have it.

    The initial V is (Vx2 + Vy2)1/2

    And if you didn't know θ, the angle is given by Vy/Vx = tanθ

    Since it turns out you knew θ, you could have simply said 4.9 = Vo*sin30 or Vo = 2*4.9

    Maybe next time post the whole problem?
     
  8. Mar 3, 2009 #7
    ah sorry i thought i put θ in there, sorry about that but thank you so much for your help, you answered a lot of questions for me that i just didnt understand.
     
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