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Suvat - Ball thrown off a roof

  1. Apr 18, 2015 #1


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    Gold Member

    1. The problem statement, all variables and given/known data
    A ball is thrown vertically upwards from a roof of a building and it lands 3 seconds later on the ground 7 meters below the roof. Calculate:
    a) the speed with which the ball was thrown upwards
    b) the maximum height of the ball above the ground
    c) the speed with which the ball hits the ground

    2. Relevant equations

    v = u + at
    s = ut + (1/2)(a)(t^2)
    s = vt - (1/2)(a)(t^2)
    v^2 = u^2 + 2as
    s = ((u + v)/2)t

    3. The attempt at a solution

    I was trying to help a friend with this problem. I have tried breaking it down to 3 sections, up, down to starting height and down to the ground; but there never seems to be enough information to solve anything. I can't tell if I am missing something obvious or something was left out of the question in the book.

    s = ? u = ? v = 0m/s a = -9.8m/s^2 t = ?


    s = ? u = 0 v = ? a = -9.8m/s^2 t = ?

    To Ground

    s = 7 u = ? v = ? a = -9.8m/s^2 t = ?
  2. jcsd
  3. Apr 18, 2015 #2
    Nothing was left out of the question in the book; the problem is quite solvable. Your breaking of the problem into three parts is a step in the right direction. Try thinking about this: If you throw a ball upwards, what must be its velocity when it comes back down to the height from which it was thrown upwards? What is the ball's velocity at its maximum height? How can you express the time it takes the ball to reach its maximum height in terms of its initial velocity and the constant acceleration it undergoes due to gravity?
    Just realized why some people call the equations for one dimensional motion with constant acceleration suvat.
    Last edited: Apr 18, 2015
  4. Apr 19, 2015 #3


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    Homework Helper

    The solution is much simpler if you use displacement (y) in terms of the time. Deciding "up" positive, the initial velocity vi is positive, the acceleration is a= -g, and the final displacement is yf= -7m. No need to break the problem.

    The equation for the displacement y=vit-g/2 t2 is valid for the whole motion. Substitute the given data for the final time and displacement, you get the initial velocity.
    After that, you can use the equation for velocity v = vi - gt, to find out the maximum height and the final velocity.
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