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Easy Angular Physics

  1. Dec 8, 2011 #1
    We just started this unit and i'm lost....

    An arrow is shot at 33◦ angle with the horizontal. It has a velocity of 55 m/s.
    The acceleration of gravity is 9.8 ms^-2
    How high will the arrow go?
    Answer in units of m

    I know that S= 1/2 at^2+ut
    and all the rules of the basic trig functions (Sin, cos, tan) I jsut don't see how to incorporate it all into this question...
     
  2. jcsd
  3. Dec 8, 2011 #2
    Draw your diagram showing the arrow and its initial launch velocity. Since you are asked how high it can go this means you are only interested in the vertical motion so you can deal only with velocity, acceleration and displacement in the y direction. See if you can go from there. There are two pieces of implied information in this problem. First, gravity is acting on the arrow, second, what happens to the velocity in the y direction when the arrow reaches the highest point?

    (There is an alternative solution using energy if you happen to think of it but it's a bit tricky)
     
  4. Dec 8, 2011 #3
    OH i think i got it now....
    okay so i used 55*sin(33) to find the y component of the velocity. then used the equation v^2= u^2+2as
    v= 0
    u= 55 m/s
    a= -9.8 m/s^2
    and using this got a height of 154.34 meters.

    Can you tell me if this is correct?
     
  5. Dec 8, 2011 #4
    scratch that. i caught my own mistake. i meant to plug in 29.96 m/s for u. and the answer i get is 91.6 meteres!
     
  6. Dec 8, 2011 #5
    you're doing it right - just double check your math. It's probably right - I don't have a calculator and am too brain dead to estimate right now.
     
  7. Dec 8, 2011 #6
    thanks! i got it right :D
     
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