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Kinematics: when to use the quadratic formula?

  1. Jul 2, 2013 #1
    Hi PF, I'm reviewing my notes from class, starting from the very beginning. I'm working on some problems using kinematic equations. Here's one example: a ball falls from 30m (using down as the positive direction) at a velocity of 8 m/s. how long does it take the ball to hit the ground?

    Why do I have to use the quadratic formula here? How do I know when not to use it?

    Thanks in advance :)
     
  2. jcsd
  3. Jul 2, 2013 #2

    mfb

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    There is no linear formula describing the motion.
    If there is no easier way to solve the problem.

    It is possible to solve that problem without solving a non-trivial quadratic equation, but I don't think this is easier.
     
  4. Jul 2, 2013 #3

    BruceW

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    Isn't it just because the acceleration is constant? hint: use integration twice on the acceleration, then you see what the equation of motion has to be.
     
  5. Jul 2, 2013 #4

    Zondrina

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    There is no horizontal motion in this case. That alone should be enough to tell you that you don't need to use it.

    You have a constant velocity and a displacement, so simply use :

    $$Δt = \frac{\vec{Δd}}{\vec{v}}$$
     
  6. Jul 2, 2013 #5

    WannabeNewton

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    Just because there is no horizontal motion doesn't mean the trajectory of the particle is not quadratic in time; I have no idea how you came to that conclusion. In the OP's case the ball undergoes free fall after being released with some initial velocity and so undergoes constant acceleration. It's trajectory will be quadratic in time.

    Just as a side note, the expression ##\frac{\Delta d}{\vec{v}}## is not well defined because you are diving by a vector (the velocity). What you mean to use is ##\left \| {\vec{v}} \right \|## i.e. the speed.
     
  7. Jul 2, 2013 #6
    This might be a silly question! But, do you mean that the ball has a speed of 8 m/s when it is at a height of 30 m?
    "Falls FROM a height of 30m" sounds to me that it was released from 30m with 0 velocity. Do you mean that or do you mean that it was thrown down with a velocity?
    Sorry if this sounds pedantic.
     
  8. Jul 2, 2013 #7

    Zondrina

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    Oh my bad, on a second read I see that using ##\vec{Δd} = \vec{v_1}Δt + (1/2) \vec{a} (Δt)^2## would be more appropriate.

    I forgot to consider the acceleration in my haste.
     
  9. Jul 2, 2013 #8

    WannabeNewton

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    It's ok! We all make silly errors from time to time :)
     
  10. Jul 2, 2013 #9
    Just as a side note, many years ago, on my analytic geometry exam, I was given a problem that I was able to reduce to ## \vec{a} \times \vec {x} = \vec{b} ##. And then I said that finding ## \vec{x} ## would require vectorial division, which is not "well defined".

    The rest was so sad I still remember that :)
     
  11. Jul 2, 2013 #10

    WannabeNewton

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    Haha. The good ol' days of analytic geometry eh? Too bad you can't make up the rules as you go along xD
     
  12. Jul 2, 2013 #11
    The problem says that initial velocity is 8 m/s, but I understand what you're saying.
     
  13. Jul 2, 2013 #12
    Haha I tried this and the result in the academic world wasn't very pretty...
     
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