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Homework Help: Two balls are thrown vertically upward, one with an initial speed twic

  1. Mar 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Two balls are thrown vertically upward, one with an initial speed twice that of the other. The ball with the greater initial speed will reach a height

    - twice that of the other
    - 4 times that of the other

    2. Relevant equations
    dy = vt + 1/2 at^2
    [if i use this equation, i will get twice of the other]

    vf^2 = v1^2 + 2ad
    -v1^2 = 2ad
    [if i use the equation, i will get 4 times of the other]

    3. The attempt at a solution

    I get both 2 times and 4 times..
    Last edited: Mar 3, 2013
  2. jcsd
  3. Mar 3, 2013 #2
    This is incorrect. You might want to show here how you get this result.
  4. Mar 3, 2013 #3
    i made a typo
    it is dy = vi t + 1/2 at^2

    this is one of the kinematic equation
  5. Mar 3, 2013 #4
    It seems to me that projectile motion is difficult for many because every second post is about it but that's ok.
  6. Mar 3, 2013 #5
    It is indeed a valid equation, but your result is still incorrect.
  7. Mar 3, 2013 #6
    Galileo created the whole science called "physics" by considering projectile motion, so it should not be surprising that it is a hot topic in teaching of this science. That's why it is so frequent here, not because it is difficult for many. Well, some problems may be surprisingly difficult.
  8. Mar 3, 2013 #7
    Yes, but in general projectile motion problems are easy because if you understand the concept you can solve the problem using only a couple of kinematic equations.
  9. Mar 3, 2013 #8

    why is it incorrect? can u explain why?
  10. Mar 3, 2013 #9
    The correct result is 4 times. I cannot explain why you are not getting the correct result using this method, you have not shown your working.
  11. Mar 3, 2013 #10

    i did not do any work. it's simply because dy - at^2 = vi t
    so there is no square sign
  12. Mar 3, 2013 #11
    What about t? Is it the same in both cases?
  13. Mar 3, 2013 #12
    no.. the ball with higher velocity has shorter time..o...i get it loll
    using vf^2 = vi^2 + 2ad , a is constant for both cases
    but if u use the other equation, t varies so the ans is not correct !
  14. Mar 3, 2013 #13
    It might be helpful for you to compute the time it takes to reach highest point in both cases.
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