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Homework Help: Speed at impact of 2 balls thrown from a building

  1. Nov 6, 2012 #1
    Hi, I'm having a bit of trouble with the below practice problem from my text book. The answer given in the back is -9.52 m/s, but there is no explanation. Thanks for any help.

    1. The problem statement, all variables and given/known data
    Two balls are thrown off of a 7.25 m high building both at an initial speed of 63.5 mph. However, one ball is thrown horizontally and the other straight down. What is the difference in the speed of the balls when they hit the ground. This is a basic problem, so things such as air resistance don't need to be taken into account.

    2. Relevant equations
    I believe that [tex]v_{y}^2 = v_{y0}^2 - 2g(y - y_{0})[/tex] is relevant here. However, that equation was introduced a few chapters before the chapter this problem is found in, so maybe I am supposed to use a different method involving Kinetic Energy, Work, or Power which are the subjects of the current chapter.

    3. The attempt at a solution
    I converted the initial speed from mph into m/s and got 28.38 m/s. I then tried to solve for [itex]v_{y}^2[/itex] in the above noted equation for each ball. For ball 1 (thrown horizontally) I used:
    [tex]v_{y}^2 = 0^2 - 2(9.81)(7.25-0)[/tex] My first problem here is that this gives me a negative sign on the right side of the equation, so I can't take the square root of both sides to solve for v. I did see an example in the book though where the negative sign mysteriously disappeared, so assuming this [itex]v = \sqrt{2*9.81*7.25} = 11.93[/itex]m/s
    For ball 2 (thrown straight down) I solved for v with the same equation:
    [tex]v_{y}^2 = 28.38^2 - 2(9.81)(7.25-0)[/tex] and got v = 25.75 m/s. The difference in these values in 13.82 m/s, which is incorrect as they should differ by 9.52 m/s according to the book.
  2. jcsd
  3. Nov 6, 2012 #2


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    Hello KarlKarlJohn. Welcome to PF !

    For ball 2:

    y = 0 ,

    y0 = 7.25 .

    You have them reversed, so you get a negative.
  4. Nov 6, 2012 #3
    Thanks for the help and the welcome! Switching those values makes sense and definitely helps with my issue with the negative, but I still don't seem to be getting the correct answer. I now have: [tex]v_{y}^2 = 0^2 - 2(9.81)(-7.25) \Rightarrow v = 11.92m/s[/tex]
    [tex]v_{y}^2 = 28.38^2 - 2(9.81)(-7.25) \Rightarrow v = 30.78 m/s[/tex]

    The difference in these velocities is 18.86 m/s, my book says the answer should be -9.52 m/s. Any idea what else I may be doing wrong?
  5. Nov 6, 2012 #4


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    What about horizontal speed?
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