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Help with a Problem Please

  1. Sep 18, 2006 #1
    Help with a Problem Please....

    This is the problem.

    A projectile is being shot from a cannon and must land in a hole 120 meter aways. What angle must the projectile be shot at to land in the hole? Here are some other numbers

    VoX=60cos(??) Voy=60sin(???)
    VfX=? Vfy=?
    Ax=0 Ay=-9.81 m/s^2
    t=?? t=??
    dx=120m dy=0
  2. jcsd
  3. Sep 18, 2006 #2
    Sorry, just read the post that I need to post homework problems elsewhere.
  4. Sep 18, 2006 #3
    take note that the time for vertical and horizontal motion is the same.

    since you know the horrizontal velocity, which is in terms of 60cosZ, can you find time?
  5. Sep 18, 2006 #4


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  6. Sep 18, 2006 #5
    I tried finding t this way dx=Vxt === 120m = 60cosZt

  7. Sep 18, 2006 #6
    yeah you are on the right track, try equate the time into your vertical motion equation.

    ( they land on the ground the same time, the 2 motions)
  8. Sep 18, 2006 #7
    The answer would be 0 as far as the vertical aspect is concerned right?? I mean the vertical displacement is 0.
  9. Sep 18, 2006 #8
    correct vertical displacement would be zero

    EDITED version
    Last edited: Sep 18, 2006
  10. Sep 18, 2006 #9
    So, would the vertical time be t = sqrt d/VoYt + 1/2a
  11. Sep 18, 2006 #10
    you can simplify the equation into 0 = t(-Voy + 1/2 g t) ( when solving this type of qns remember to state your sign convention, you don't want the case where you get negative time)

    so Voy = 1/2 g t
    Last edited: Sep 18, 2006
  12. Sep 18, 2006 #11
    So t=VoY/2g. But how do I find that if I do not know the angle?? You cant divide 60sin of nothing by 2g can you?
  13. Sep 18, 2006 #12
    you can equate the 2 equations u have together, the horizontal and vertical, remember the time is the same for both.
  14. Sep 18, 2006 #13
    Equate them together? So.... t= (120/60cosZ) * (60sinZ/2g)
  15. Sep 18, 2006 #14
    if you equate in such a way, you have 2 viarables to solve and only one equation to work with.

    BTW your equation should be T^2 = ......, since u times the 2 equation together.( but this is not the correct equation to use.)
    Last edited: Sep 18, 2006
  16. Sep 18, 2006 #15
    So that is not the correct way to equate it?? I am really confused here.
  17. Sep 18, 2006 #16
    T = 120/ (60 cos Z)

    T = 2(60 sin Z)/(9.81)

    120/(60 cos Z) = 2(60 sin Z)/(9.81)

    EDIT : i amended the equations. For the time derieved by vertical motion,

    Voy = 1/2 g t
    (2 voy)/g = t

    t = (2 (60) sin Z) / g

    You wrote wrongly the equation for your previous post.
    Last edited: Sep 18, 2006
  18. Sep 18, 2006 #17
    Ok. I'm sorry. I thought I put a equals between the two equations. I did not realize it was a muliplication sign. What do I solve for now?? Do I set the equation equal to 0 and divide -120 by 2(9.81)?
  19. Sep 18, 2006 #18
    you can cross mutiply. Then isolate your Z, i used double angle formular for this.

    BTW this is the first time i use double angle formular for physic. feels strange
  20. Sep 18, 2006 #19
    So, 120 m * (2*9.81) = 2354s^2. What would 60 sinZ time 60cosZ be???? 3600 sinZ????
  21. Sep 18, 2006 #20
    this part is wrong, as i have stated about, the equation is

    120/(60cos Z) = ((2)(60)sin Z)/ 9.81

    thus the correct expression should be 120(9.81) = (2)(60)(60)cos Z sin Z
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