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Projectile motion

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown at the angle 30 degrees with an initial speed of 6 m/s, it takes the ball 3 seconds to hit the ground. Calculate the height where the ball was thrown
    [​IMG][​IMG]

    2. Relevant equations



    3. The attempt at a solution

    I tried to solve this using the equation:

    yf = yi + Viy *t - 0.5 * g * t^2
    yf-yi = (6 * sin(30) * 3 s) - (0.5*10*3^2)
    = 9 - 45
    = -37

    why is it negative? What am I doing wrong here?
     
  2. jcsd
  3. Oct 1, 2008 #2

    LowlyPion

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    The number is negative because it lands below the point it was thrown.
     
  4. Oct 1, 2008 #3
    so the answer is right 37 ? or is it supposed to be 9+45?
     
  5. Oct 1, 2008 #4

    LowlyPion

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    It would be 9 + 45 if you threw it downward at the same speed and angle to the horizon.

    By the way the correct answer for 45 - 9 is 36 and not 37.
     
  6. Oct 1, 2008 #5
    ok, then to find the final velocity when the ball hits the ground I need to find:

    Vyf^2 = Vyi^2 -2gh
    = (6* sin 30)^2 - 2*(10)*(45)
    = 9 - 900
    = sqrt (-891) ??
     
  7. Oct 1, 2008 #6

    LowlyPion

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    That would be because that is not a correct equation. For instance are you saying that if initial velocity was 30 m/s upwards, that by your equation that final velocity would be 0?

    To use that equation you would need to find the maximum height first, because what you are using describes a velocity translation in one direction.
     
  8. Oct 1, 2008 #7
    so what equation should I use here?
     
  9. Oct 1, 2008 #8

    LowlyPion

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    Was that asked as part of the question?

    Because if it was you have to use both components of velocity.
     
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