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Throwing a ball from the roof of a building

  1. Dec 1, 2005 #1
    A ball is thrown horizontally from the roof of a building 54 m tall and lands 50 m from the base. What was the ball's initial speed?

    I have been set this question and I'm a little stuck...here is the information I have extracted from the question:

    sv = -54
    sh = 50
    v = ?
    a = -9.81

    I don't know what else I can do, or what formulas/principles I should be using. I can't modulate with Pythagoras as surely the ball wouldn't travel downwards in a straight line to it's resting point...any help?

    Much appreciated,
  2. jcsd
  3. Dec 1, 2005 #2
    (forenote, an underscore indicates a subscript)

    You're familiar with the basic equation of motion relating distance to velocity and time, correct?
    [tex]d = d_o + V_o*t+\frac{1}{2}a*t^2[/tex]
    d_o = original distance
    V_o = original velocity
    a = acceleration (in our case, gravity)
    t = time

    You have two unknown variables: Original Velocity and Time
    Well, here's what you'd do.
    V_o_x = V_o
    V_o_y = 0

    Because the ball has no initial vertical velocity component, you can solve for the free fall equation [tex]d_f_a_l_l = 0.5g*t^2[/tex]
    solving symbolicaly, we obtain [tex]t = \sqrt{\frac{2*d_f_a_l_l}{g}}[/tex]
    ** Substitute your building height and g (9.81 m/s^2) into this equation and solve for time

    Now you can use the second motion equation Distance = Velocity * Time to acheive the original horizontal velocity.

    Look at it this way; the ball travels a known distance horizontally, and it does it in a known amount of time, because we know how long it would take for the ball to fall. Because there is no horizontal acceleration, we can solve the horizontal velocity equation relatively easily simply by setting:
    [tex]V_o = d_h / t_f_a_l_l[/tex]
    which gives us
    [tex]V_o = \frac{d_h}{\sqrt{\frac{2*d_f_a_l_l}{g}}}[/tex]

    I know it looks nasty, but it's really quite easy once you start getting some solid numeric values. I'll leave the calculations up to you.
  4. Dec 1, 2005 #3
    Wow thanks for that. Ok so this is what I've done now:

    t = sqrt [ (2 * d_fallen) / g ]
    t = sqrt [ (2 * 54) / 9.81 ]
    t = sqrt [ 108 / 9.81]
    t = sqrt 11.009
    t = 3.318

    d_horizontal = vt
    v = s / t
    v = 50 / 3.318
    v = 15.069

    Does this seem like the correct answer? Cos I only have one submission for this question, and I really don't wanna get it wrong, I wanna know how to answer questions like this! So does 15.069 ms^-1 seem about right? Or have I gone wrong somewhere in my calculations...
  5. Dec 1, 2005 #4
    Approximation, estimation, generalization

    if you're unsure about your answer, you can do one of two things to check it...
    A) Solve it backwards using your answer.
    B) Graph it. It's easy, really... just set X as the time variable and d_y as the y variable, and you should easily be able to see if your answer for time @ y=0 is correct...
    Learn to love your TI-83, 83+, 86, 89... :approve:

    Attached Files:

  6. Dec 2, 2005 #5
    ah thanks a lot i got the answer correct, much appreciated.
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