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Question on Projectile

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data
    The ball is now thrown with the same speed at an angel of 35 deg upward. How far does it travel?

    My question is...

    They have

    vhor = (25 m/s) cos 35 deg = 20.5 m/s

    and

    The angel with the vertical is (90 - 35) = 55 deg
    so...

    vvert = (25 m/s) cos 55 deg = 14.3

    It seems backwards to me for some reason because it says that the ball was thrown at a angel of 35 deg upward.

    Can someone help me understand why 35 deg is tacked onto the horizontal while 55 is tacked onto the vertical?
     
  2. jcsd
  3. Feb 18, 2009 #2
    Usually when talking about projectile motion the angel is to the horizontal line. So you aim 35 degrees up from the horizon. And if you take a closer look at the geometry of a triangle you'll see it is possible to use the same angel for horizontal and vertical speed. Just keep in mind that you can move vectors around.
     
  4. Feb 18, 2009 #3
    Ok I do not get how the angel with the horizontal is 35 while the vertical is 55 because you aimed 35 degree upward.
     
  5. Feb 18, 2009 #4
    I hastely put together a picture of the situation.
    http://www.aijaa.com/img/b/00677/3628783.jpg [Broken]
    a is the vertical speed, b is horizontal speed and c is a and b combined. Those two pictures mean the exact same thing, I've just moved a. How would you solve a from the picture on the right, if you only knew the value of c and the angel?
     
    Last edited by a moderator: May 4, 2017
  6. Feb 18, 2009 #5
    Thank you for the visual. I am new to the whole SOH, CAH, TOA and that visual helps greatly...

    C = 25m/s
    B = 20.5m/s
    A = 14.3
     
  7. Feb 18, 2009 #6
    So did you find out, why you can also use 35 degrees instead of 55? Can you show me how you got those figures?
     
  8. Feb 18, 2009 #7
    (35 sin) 25 = 14.3

    They did it by subtracting 35 from 90 and doing (55 cos)25 = 14.3
     
  9. Feb 18, 2009 #8
    Yep!
     
  10. Feb 18, 2009 #9
    Thank ya for helping me figure it out.
     
  11. Feb 18, 2009 #10
    You're very welcome!
     
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