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Projectile Question Physics 12

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi, I am just wondering if someone can check if this is the correct answer for this question. Thanks in advance

    1 . A projectile is launched with a speed of 48m/s at an angle of 34 degrees BELOW the horizontal. Calculate the speed an direction of the projectile 2.5 seconds after launch. Ignore friction.


    2. Relevant equations

    x/y component . addition of vectors. v=vo+at

    3. The attempt at a solution


    This is my answer : 40 m/s at 3.3 degress South of East

    I find it surprising that the final velocity of the projectile is less then the initial (which is 48m/s).


    Here it is:

    48 sin 34 = -26.8m/s
    48 cos 34 = 39.8

    v = vo +at

    v = -26.8 + (9.8)(2.5) <--- + acceleration because motion is towards center of Earth

    v = -2.3

    From here I used addition of vectors in x/y component

    x = 39.8
    y = -2.2

    After 2.5 seconds

    Answer: vf = 40 m/s at an angle of 3.3 degrees <---- tan theta


    Shouldn't the vf of the projectile (after 2.5 seconds) be greater then the vo? Since it has a + 9.8 m/s ^ 2 .

    Thank you
     
  2. jcsd
  3. Mar 28, 2009 #2

    Delphi51

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    Homework Helper

    You need a minus sign on the 9.8. The 26.8 and the 9.8*2.5 are both downward.
    For the direction, you'll get an angle below horizontal - no "South of East"
     
  4. Mar 28, 2009 #3
    Thank you very much. The correct answer should be 65 m/s , 55 degrees below horizontal ?
     
  5. Mar 28, 2009 #4

    Delphi51

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    Homework Helper

    I got 64 m/s and 51 degrees, but I used g = 9.81.
    Might be worth checking the angle again - that's quite a difference.
    Did you have 49.7 for the vertical velocity?
     
  6. Mar 28, 2009 #5
    Nope, i got 51.4 m/s...

    But.... it was 52 degrees. I dont know why i put down 55 degrees lol

    Thanks
     
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