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Projectile motion

  1. Jan 15, 2009 #1
    Ok, I thought I knew how to do this, but for some reason, according to my homework website, my answer is wrong. So I'm turning to this board before I beat my head against the wall. Please help me.

    1. The problem statement, all variables and given/known data
    A ball is thrown with an initial speed of 31 m/s at an angle of 45�. The ball is thrown from a height of 12 m and lands on the ground.
    Find the time of flight.

    3. The attempt at a solution
    Initial vertical upward velocity = 31 sin 45 = 26.378m/s.
    The displacement [Yf – Yo] = –12 m
    Yf -Yo = Vot - 0.5*g*t^2
    –12 = 26.378 t – 4.905 *t^2
    4.905 *t^2 – 26.378 t – 12 = 0
    --------------------------------------...
    Using b^2 – 4ac = 26.378 ^2– 4*4.905*(– 12)
    √ (b^2 – 4ac) = 30.516
    – b ±√ (b^2 – 4ac) = 26.378 ± 30.516
    = 56.894 leaving the negative time value
    Dividing by 2a = 9.81
    t = 5.7996 s

    Is my work right?
     
  2. jcsd
  3. Jan 15, 2009 #2

    LowlyPion

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    Homework Helper

    Welcome to PF.

    I'd check the vertical velocity component again.

    31 * sin45° = ... ?
     
  4. Jan 15, 2009 #3
    Ugh, no wonder none of my answers have been right, I've been in radian mode on my calculator. I feel like an idiot. Thank you for pointing out the obvious to me.
     
  5. Jan 15, 2009 #4

    LowlyPion

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