Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the speed of a ball given some height and falling speed

  1. Aug 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Nicole throws a ball straight up. Chad watches the ball from a window 6.10 m above the point where Nicole released it. The ball passes Chad on the way up, and it has a speed of 11.0 m/s as it passes him on the way back down. How fast did Nicole throw the ball?

    2. Relevant equations

    x= x0 + v0t + 1/2 at^2
    v= v0+at

    3. The attempt at a solution

    I haven't the slightest clue as how to solve this. If she's tossing it upward, I know that would be a parabola graphically, so perhaps I need to use the quadratic equation at some point.
    Need to find acceleration, I suppose, to find how fast she threw the ball?
  2. jcsd
  3. Aug 25, 2011 #2


    User Avatar
    Homework Helper

    The acceleration involved is the acceleration due to gravity.
    If the ball was doing 11 m/s on its way down past the window, it will have been travelling at 11 m/s on its way up past the window as well.

    Be careful about which figures are positive and negative when calculating.
  4. Aug 26, 2011 #3
    So...the answer is just 11 then? It was a trick question?
  5. Aug 26, 2011 #4


    User Avatar
    Homework Helper

    NO, it had slowed to 11 m/s by the time it went past the window, it must have been going faster when it left her hand.

  6. Aug 26, 2011 #5
    So it slows down from some unknown number due to gravity. Which would mean negative acceleration. But then how do you figure out the unknown, and how does 11 work into the equation?
  7. Aug 26, 2011 #6


    User Avatar
    Homework Helper

    When I do these problems I use the following symbols

    v = final velocity
    u = initial velocity
    s = displacement
    a = acceleration
    t = time

    For the ball, we know it had slowed to 11 m/s by the time it got to the window 6.1 m up

    v = 11
    a = -9.8
    s = 6.1

    You want to know the initial speed , "u" , so the formula of choice would be

    v^2 = u^2 + 2as [It is the only one of the 5 that does not require a time value]

    121 = u^2 + 2*-9.8*6.1

    you should be able to do it from there. Be careful with that minus sign.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook