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Projectile Motion analyzed

  1. May 13, 2010 #1
    1. The problem statement, all variables and given/known data
    In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the spike is difficult.Suppose a ball is spiked from a height of 2.30 m with an initial speed of 20.0 m/s at a downward angle of 18.00o. How much farther on the opposite floor would it have landed if the downward angle were , instead, 8.00o?
    (Source - Fundamentals of PHYSICS, Halliday, Resnick,Walker, 8th edition, chapter 4, QS.33)


    2. Relevant equations
    This is a case of projectile motion , thus,
    say, initial velocity = u,
    displacement in vertical direction = h
    displacement in horizontal direction = s
    time of flight = t

    h = uyt + 0.5ayt2
    s = uxt + 0.5axt2


    3. The attempt at a solution

    CASE 1:angle = 18o
    uy = usin(18o) = 6.18m/s

    -2.3 = 6.18t -4.9t2
    t = 1.562s

    ux= ucos(18o) = 19.02 m/s
    so, s1 = 19.02 * 1.562 = 29.71 m

    CASE 2:angle = 8o
    uy = usin(8o) = 2.78m/s

    -2.3 = 2.78t-4.9t2
    t = 1.025s

    ux = ucos(8o) = 19.81m/s
    so, s2= 19.81*1.025 = 20.31m:eek:

    Well, I am confused.:confused:

    Thank you.
     
  2. jcsd
  3. May 13, 2010 #2

    rl.bhat

    User Avatar
    Homework Helper

    -2.3 = 6.18t -4.9t^2

    h, uy and ay are all in the same direction. So all should be either positive or negative.
     
  4. May 14, 2010 #3
    Thank you so much! That points out the importance of signs and vector diagrams.
    Yes, so the equations are:
    CASE 1: angle = 18o
    -2.3 = -6.18t -4.9t2
    t = 0.3s

    ux = ucos(18o) = 19.02m/s
    so, s1 = 19.02 * 0.3 = 5.706 m

    CASE 2: angle = 8o
    -2.3 = -2.78t - 4.9t2
    t = 0.46s

    ux = ucos(8o) = 19.81m/s
    so, s2 = 19.81 * 0.46 = 9.113 m

    Hence , difference = s2-s1 = 9.113 - 5.706 = 3.407 m.

    ANS = 3.407m:approve:

    Thank you!:cool:
     
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