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Quick question (not homework) just wondering

  1. Aug 28, 2013 #1
    If I throw a ball vertically in the air and catch it again at the exact same height after 3.6 seconds. (looking away from air resistance)

    So the question is: 1. How big is the starting speed? (v0) And how far in the air would the ball go?

    2. If we were not to catch the ball it would've hit the floor after 3.9 seconds after we threw it. How far away from the floor is that starting point?

    I think if you put v = 0 (when it's at its highest point) and you got g ofc (9.81m/s) But I really can't figure it out.. which formulas to use etc, I feel like I have too little information? But if I set v = 0 and calculate the time and the distance to the highest point I think I can solve it... only problem is that I don't know how to calculate the first part of the problem.
     
  2. jcsd
  3. Aug 28, 2013 #2

    CAF123

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    You should be able to find out the time taken for the ball to reach the apex of its trajectory. Then, as you said, it's velocity there is zero so you can solve for vo using a kinematic equation.

    Even though this may not be h/w, I believe it still belongs in that section.
     
  4. Aug 28, 2013 #3
    What do you mean? kinematic equation? You mean the Ek =1/2mv^2 and 1/2mv0^2 and the Ep=mgh stuff?

    Couldn't figure out how to type the mv thingy with the start speed v0 and also put it as ^

    Actually I didn't pay attention in the first physics I had so I am very bad on these formulas. Never really learnt it, I have to learn it now! Can you show me or something?
     
  5. Aug 28, 2013 #4

    jtbell

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    Yes, even though this is not actually a homework question, it is a homework-type question (asking for help in solving a specific exercise) and therefore belongs in the homework forums.

    Because you've already gotten some help, I've move this thread for you. In the future please use the correct forum for questions like this. Read the "How to ask for help" post that is pinned at the top of this (new) forum, and fill out all sections of the template that is provided when you start a new thread in this forum.

    (and welcome to Physics Forums!)
     
  6. Aug 28, 2013 #5

    haruspex

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    There is a collection of useful standard kinematic equations that apply when acceleration is constant. They can be derived easily enough by calculus, but it's worth learning them.
    Using s = distance traversed, a = acceleration, u = initial speed, v= final speed (many use vi, vf for these), t for elapsed time:
    v = u + at (obvious... increase in speed = accn * time)
    s = ut + at2/2 (integrate eqn above)
    v2 - u2 = 2as (this is the same as the conservation of work you hinted at, but with the m cancelled out)
    s/t = (u+v)/2 (two expressions for average speed)
    Note that each references a different four of the five variables. So to pick which equation to use, just see which three variables you are given and which you want to determine.
     
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