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Kinematics in Two Deminsions, Projectile Motion

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data
    I have the angle and intial velocity but i need to find distance traveled in the x direction (i think).

    In the Olympic shotput event, an athlete throws the shot with an initial speed of 12 m/s at a 40.0 (DEGREE) angle from the horizontal. The shot leaves her hand at a height of 1.8 m above the ground.

    Known:
    Angle (in degrees): 40
    y inital: 1.8 m
    Initial Velocity: 12 m/s
    Intial Y compontent for velocity: 12sin(40) ?

    2. Relevant equations
    y= yo + voy – (0.5)g(t^2)


    x = Vo(cos(40))t ?????


    3. The attempt at a solution

    0 = 1.8m + 12(sin40)t + 1/2g(t^2)
    Solve for zeros of t?
    The only realistic one comes out to be about 1.7 since the other number is negative.
    x = 12cos(40)t (plug t in) and get: 15.6m ????????

    The correct answer seems to be 16.6m so i know i must doing something wrong.
     
  2. jcsd
  3. Mar 3, 2009 #2
    You've set it up the way I would set it up, as far as I can tell. I don't have my calculator on me, so I'm not checking for silly math errors, but unless I'm missing something too, the set-up looks correct.

    I've got an exam on this stuff Thursday, I hope I'm right in saying you set it up correctly.
     
  4. Mar 3, 2009 #3

    LowlyPion

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    Homework Helper

    I think you need to consider g is (-). But apparently you did to get 1.7 (Actually I get 1.782) If you had used +g you would have gotten 2 (-) times.

    Perhaps if you carried greater precision through your calculation you would get a better answer?

    Otherwise your method is fine.
     
  5. Mar 3, 2009 #4
    I basically just divide g by 1/2 and end up entering it (-4.9)

    Also, when i take the 1.782 and put it into the second equation it doesn't equal 16.6?
     
    Last edited: Mar 3, 2009
  6. Mar 3, 2009 #5
    Nevermind. The online system i am entering in the answers for is picking up on my round errors. Luck Me. =)
     
  7. Mar 3, 2009 #6

    LowlyPion

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    Yes my point was that your equation showed a + sign instead of -g. As long as you keep it straight is what's important. Just be careful.

    I also meant that you would use Sin and Cos to greater precision.

    I see that you got it, and what's more important is that you understand how you got it, so ... all's well then.
     
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