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Help with Physics Lab

  1. Oct 17, 2003 #1
    Hey guys, thanks for reading my post. I'm really confused about a lab that was assigned by my teacher. I will try to explain the whole lab the best I can.

    Lab: The Balls Lab

    Purpose of the Lab: To determine the initial velocity (Vo) and maximum height (dy) of a thrown (and hit) ball

    Materials: tennis ball, tennis racket, stop watch
    1. Throw a ball straight up into the air
    2. Measure the time the ball is in the air and comes down
    3. Due three trials

    Data: (here are the 3 time results I got for the ball)
    Trial 1- 3.21 seconds
    Trial 2- 3.10 seconds
    Trial 3- 2.94 seconds
    Average- 3.08 seconds

    Hitting the ball with tennis racket
    Trial 1- 1.41 seconds
    Trial 2- 2.88 seconds
    Trial 3- 1.87 seconds
    Average- 2.05 seconds

    Ok here is the problem... I have no clue how to start solving for the initial velocity (Vo) for each of the average result of the ball thrown and hit in the air. At first I thought the initial velocity of the thrown ball was 0 because that is usually the velocity of any object at rest and then going into the air. But my teacher said it wasn't because he wanted the intial velocity from when it left my hand not when it is at rest, which is 0. He also told me that I needed to use one of the equations of motion for constant acceleration such as: Vf=Vo+at, d=1\2(Vf+Vo)t, d=Vo=1/2at(squared), or 2ad=(Vf)squared-(Vo)squared and the acceleration of gravity, -10 m/s to solve for Vo. But he wasn't clear on which one. Also it seems to me that each of these equation require the value of Vo itself! I am also very confused at how to find the height of the thrown ball (dy) because y is not given as a variable in any of these equations.

    If you guys can show me how I can solve for the initial velocity (Vo) and the maximum height(dy) of the of the ball thrown using any of the equations above or even your own methods, please enlighten me with you explanation. I am most grateful.

    Thank you for taking the time to answer.
    Last edited by a moderator: Oct 17, 2003
  2. jcsd
  3. Oct 23, 2003 #2


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    Science Advisor
    Gold Member

    Your procedure should have included stopping the stopwatch when the ball comes down to the exact (or as close as you can get it) position that the ball was in before it was thrown. If you did that, then everything else is no problem. The time to go up is the same as the time to go down, so if you divide the total flight time by 2, you'll have the time it takes to get to the maximum height. You also know that when it gets to the top, its velocity has been decelerated to 0. So use v = vo + a*t here to solve for vo (plug in v = 0, and a = -9.81m/s/s, and t from your experiment divided by 2). To get the max height from there, just use y = .5*a*t^2 + vo*t. You just calculated vo, t is again the time from your experiment divided by 2, and a is -9.81m/s/s again.
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