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Projectile: time max height reached

  1. Jun 21, 2011 #1
    1. The problem statement, all variables and given/known data
    "A boy throws a rock with an initial velocity of 3.13 m/s at 30.0 degrees above the horizontal. How long does it take for the rock to reach the maximum height of its trajectory?"

    vi = 3.13 m/s
    tmax height = ?
    vf = 0 m/s (at max height)
    a = -9.8 m/s2


    2. Relevant equations
    vf = vi + at

    (Not sure!)




    3. The attempt at a solution
    vf = vi + at
    t = [(vf - vi) / a] = [(0 m/s - 3.13 m/s)/-9.8 m/s2]
    t = 0.319 s

    Concern:
    The correct answer is listed as 0.313 s; because my answer is so close, I wondered if this were a typo. If not, I clearly am taking the wrong approach and would appreciate some guidance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 21, 2011 #2

    ehild

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    The ball has been thrown at an angle, and the magnitude of the initial velocity is 3.13 m/s. With what initial velocity does the ball move upward?

    ehild
     
  4. Jun 21, 2011 #3
    Hmmm . . . so, I should determine viy? If so, 3.13 m/s sin(30) = 1.57 m/s . . . sorry--not clear on how this will help me arrive at what's noted as the correct answer. I did plug this figure into vf2 - vi2 = 2ad, and then determined d from that . . . and then plugged that d into d = vit + 1/2at2 to determine t . . . but I ended up with the same answer, 0.319 s.
     
  5. Jun 21, 2011 #4

    ehild

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    You have the formula vf=vi+at already. Apply it to the vertical velocity components. At the maximum height, vy=0 (not the velocity v, as the ball keeps it horizontal velocity component during the whole flight)

    ehild
     
  6. Jun 21, 2011 #5
    Lol, this is going to seem so simple after . . .

    You used: -9.8m/s2

    They used: -10m/s2

    You got the correct answer under a more precise measurement.
     
    Last edited: Jun 21, 2011
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