physics

A person throws a physics book off of a cliff at an initial speed of 3 m/s at an angle of 20 degrees. The book is in
the air for a total of 2.5 seconds.
a) How high is the cliff? b) How far away from the base of the cliff does the book land?
c) What is the magnitude and direction of the book's velocity when it lands?
d) Discuss whether or not your answers to parts a) through c) are reasonable. Give specific reasons why you
think they are or are not reasonable.

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asked by ryan
  1. break the initial velocity into vertical and horizontal components.

    Vertical:
    hf=0=hi + Viv*t-1/2 g t^2 solve for hi
    final vertical veloicty: Viv-g t
    horizontal: Vf=Vih

    magnitude of velocity:
    sqrt(Vfh^2+ Vfv^2)

  2. (a) the motion of the book is projectile, and uniformly accelerated motion. thus we use the formula,
    h = ho + vo*t - (1/2)*g*t^2
    where
    h = final height
    ho = initial height
    vo,y = initial vertical velocity = vo*sin(theta)
    t = time
    g = acceleration due to gravity (9.8 m/s^2)
    making the ground as the reference or the origin (thus at this point, the height = h = 0), we solve now for ho:
    0 = ho + vo,y*t - (1/2)*g*t^2
    0 = ho + 3*sin(20)*2.5 - (1/2)(9.8)(2.5)^2
    0 = ho - 28.06
    ho = 28.06 m

    (b) since horizontal velocity, v,x is always constant, initial horizontal velocity = final horizontal velocity, or
    vo,x = v,x
    but vo,x = vo*cos(theta) , thus
    v,x = 3*cos(20)
    v,x = 2.82 m/s
    we can use this to get the range (horizontal distance):
    R = v,x * t
    R = 2.82*2.5
    R = 7.05 m

    (c) note that we already solved for the final horizontal velocity, v,x = 2.82 m/s (positive/to the right)
    we solve now for the final vertical velocity using
    vf,y = vo,y - g*t
    substituting,
    vf,y = 3*sin(20) - 9.8*2.5
    vf,y = -23.5 m/s (negative/downward direction)
    getting direction,
    arctan(23.5/2.82) = 83.1 degrees

    hope this helps~ :)

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    posted by Jai

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