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A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an angle of 36.9° above the horizontal. You can ignore air resistance. (a) At what two times is the base- ball at a height of 10.0 m above the point at which it left the bat? (b) Calculate the horizontal and vertical components of the baseball’s velocity at each of the two times you found in part (a). (c) What are the magnitude and direction of the base- ball’s velocity when it returns to the level at which it left the bat

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2 answers
  1. vertical problem
    Vi = 30 sin 36.9 = 18 m/s upward at t = 0
    v = Vi - 9.81 t
    h = Vi t - 4.9 t^2
    so for part a
    10 = 18 t - 4.9 t^2
    solve quadratic for t one and t two
    when you have t
    use
    v = 18 - 9.81 t
    for vertical component of velocity
    u = horizontal velocity = same forever = 30 cos 36.9

    part c is a trick question
    u is the same forever
    by conservation of energy or just looking at the parabola, v down at finish = v up at start :)

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    Damon
  2. Vo = 30m/s[36.9o]
    Yo = 30*sin36.9 = 18 m/s.

    a. Y^2 = Yo^2 + 2g*h = 18^2 + (-19.6)10 = 128
    Y = 11.3 m/s. at 10m.
    Y = Yo + g*T1 = 11.3
    18 + (-9.8)T1 = 11.3
    T1 = 0.684 s.

    Y = Yo + g*T2 = 11.3
    0 + 9.8T2 = 11.3
    T2 = 1.15 s.

    b. X = Xo = 30*cos36.9 = 24 m/s.
    Y = Yo + g*T1 = 18 + (-9.8)0.684 = 11.3 m /s.
    Y = 0 + 9.8*1.15 = 11.3 m/s.

    c. 30m/s[36.9o].

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