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Kinematics: Free fall problem.

  1. Mar 16, 2017 #1
    1. The problem statement, all variables and given/known data
    "A ball is thrown vertically upward with a speed 'v' from a height 'h' metre above the ground. The time taken for the ball to hit the ground is."

    2. Relevant equations

    ##s = ut + \frac{1}{2}at^2##

    3. The attempt at a solution
    https://scontent.fdel3-1.fna.fbcdn.net/v/t1.0-9/17352002_706036796237108_6058839347900308786_n.jpg?oh=3f0a769b256f51becc5ee351dc25294b&oe=596F1319 [Broken]
    https://scontent.fdel3-1.fna.fbcdn.net/v/t1.0-9/17264650_706036792903775_4158276045728207185_n.jpg?oh=ab150cc42047d1d745b8b84a53fe094e&oe=596262E0 [Broken]
    So we throw the ball vertically upward with a velocity 'v' from a reference point 'o' as I have labelled in the diagram. The ball returns back to the reference point after an interval of t1, and the displacement in this time period is 0. I used the equation ##s = ut + \frac{1}{2}at^2## and solved the quadratic equation for t1.

    Now the ball is back at o, but this time it has velocity -v, and is falling down. I assumed it took the ball time t2 to fall from o to h. Again using the equation ##s = ut + \frac{1}{2}at^2## and solving the quadratic equation for t2, I calculated t2.

    Then I added t1 and t2 for the total time taken for the fall to reach the ground. But my answer doesn't match the answer given in book. The answer given in the book is ##t = \frac{v}{g}[1+\sqrt{1+\frac{2gh}{v^2}}]## . I wonder what did I get wrong. Thank you for your help.
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Mar 16, 2017 #2

    BvU

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    I wonder too. Show your steps, so we can try to pinpoint it...
     
  4. Mar 16, 2017 #3
    You can work it out in one go without having to analyze two parts of the motion. Think about it. For example, define the ground as the origin for the position, so now you have an initial position.
    $$y(t) = h + vt - \frac{1}{2}gt^2$$
    Do you know what you need to do now to solve for the required time?
     
  5. Mar 16, 2017 #4

    BvU

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    Although: do you think the results are very different ?
     
  6. Mar 16, 2017 #5

    BvU

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    And: you want to get rid of the ##\pm## sign: there is only one answer...
     
  7. Mar 16, 2017 #6
    You can check my steps, I have uploaded photographs.
    I am such an idiot, I forgot to manipulate my answer to match the answer given in the book when my answer does indeed match with the one given in book. I apologise, for the unnecessary bother. Thank you.
    Anyway, how do I know that I have to remove ##\pm## sign?
     
    Last edited: Mar 16, 2017
  8. Mar 16, 2017 #7

    andrevdh

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    You could also use your original equation with s = -h :wink: to solve for t
     
  9. Mar 16, 2017 #8
    You need to see which makes sense physically. If a specific sign ends up giving you a negative answer, when time can't be negative (initial value problems like this one mean the time starts at 0 and cannot be anything less), you can discard that sign.
     
  10. Mar 16, 2017 #9
    I see. Thanks. Can I remove this thread, or is it a policy of physics forums to keep all threads, no matter how idiotic they are? (Like mine, as I already had the answer).
     
  11. Mar 16, 2017 #10
    Leave it here. fwiw, your thread was not idiotic at all, no question ever is. Look at it this way, andrevdh and me also gave you a quicker way to arrive at your answer, which may prove useful to you in the future, and you also learned what to check to see when you can remove a sign from the solution of a quadratic.
     
  12. Mar 16, 2017 #11
    That surely reduces all the work I had to do to solve the problem. I didn't know the original equation was powerful enough to give me answer in one go. I am new to physics, so I don't have much experience. I couldn't imagine in my head how the above equation could account for the time interval of the whole motion.
     
  13. Mar 16, 2017 #12
    In essence, as long as the ball is subject exclusively to the uniform acceleration of gravity, the equations apply. The moment it hits the ground, or if there's a gust of wind, or if anything changes the initial hypotheses (ball subject exclusively to uniform gravity) then the equation is no longer valid. In your example, from the initial moment right up to when it hits the ground, the equation holds and it describes the motion of the ball.
     
  14. Mar 16, 2017 #13

    andrevdh

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    Don't worry I also did not realize it when I started out, but it helps if you do not just jump in. Try to think if there could be other ways to solve the problem.
     
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