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Parabolic path of ball-speed before hitting the ground

  1. Dec 22, 2013 #1
    1. The problem statement, all variables and given/known data

    You toss a ball straight up at 7.3ms^-1; it leaves your hand at 1.5m above the floor.
    a) Find its velocity before it hits the floor
    b) Suppose ball is tossed straight down at 7.3ms^-1, what is the final velocity just before it hits the floor.
    c) When would the second ball hit the floor.

    2. Relevant equations

    None

    3. The attempt at a solution

    Capture.JPG
     
  2. jcsd
  3. Dec 22, 2013 #2

    Curious3141

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    Don't have time for a long post tonight, but it just seems to me that you're doing a lot of unnecessary work here - and it's also easy to make mistakes when overworking a problem.

    You've posted a lot of kinematic problems, and I've helped you with some of them. I wanted to say something before, but I'll say it now: in kinematics, it's always good to start by writing down the principal variables, and then listing the ones you're given, then finally what you need to determine. I'm not sure what form of equations you've learnt, but this is what I know:

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

    Those are the three you *have* to remember.

    Then there are these two which are easily derived from the others, but they're good to remember as well:

    ##s = vt - \frac{1}{2}at^2##
    ##s = \frac{1}{2}(u+v)(t)##

    The variables are a,u,v,t and s. Start by listing those out and writing what you know.

    For instance, in this question's part a), you're given u (=+7.3),a (=-g) and s (=-1.5). You're asked to determine v. Which equation would you select for a one step solution? Proceed like that.

    Would you expect the answers to parts a) and b) to be different? Even without working them out?
     
  4. Dec 22, 2013 #3
    Vf^2 - vi^2 =delta y

    I know man, been trying to get a head start or next semester's unit during this summer break. Just finished foundation year in uni.
    There's a total of 86 questions. I've finished 83 question. Will move on to the next topic when done.

    EDIT: Same because vi are both the same weather the ball is thrown straight up from ground of if the ball is tossed at the same speed as if it were thrown straight up from ground.
     
    Last edited: Dec 22, 2013
  5. Dec 22, 2013 #4

    Curious3141

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    No worries, you're doing quite well. But my gentle suggestion is to try and be a little more systematic, as I mentioned in my previous post. This question doesn't actually need you to solve a quadratic at all, if you work through it in the order given.

    I guess you meant ##v_f^2 - v_i^2 = 2a\Delta{y}##, which is the form you learnt it in? That's fine, and if you apply it properly there shouldn't be an issue.

    Remember what I said before about sign conventions? Easiest to let "up" be positive and "down" be negative, because that's intuitive to most people.

    So in part a), what's ##v_i##?

    And in part b), what's ##v_i##?

    What are ##a## and ##\Delta{y}## in each case? Are they the same?

    So what is ##v_f## in parts a) and b)? The same or different?

    Remember that when you're given a constant acceleration, the kinematic equations can always be applied directly. Even if the trajectory involves going up (because of an initial upward throw), then down (because of gravity acting downward), you don't actually have to calculate the maximum height, time taken to reach that, etc. unless you're actually asked those things. Immediately after leaving the hand, the only force that acts on the object is gravity, and that's constant, so the kinematic equations apply to the whole trajectory. No need to break it up into an 'up' and 'down' stage. Just be careful with your signs and interpret any times you get intelligently, and you should be fine.
     
  6. Dec 22, 2013 #5
    I just edited a few seconds before your new post.
    vi in both circumstances are both the same whether ball is thrown up at 7ms^-1 or thrown down at 7ms^-1.

    My mathematical workings reveals the above is true: vf = 9.09ms^-1
    Yes and I did allowed g to be negative.
    vf^2 - vi^2 = 2(-g)(delta y)
    I think I see where the confusion came from. I've always been using negative g but in the event where a numerical g is not required, I would write-suppose any kinematics equation with a g variable as vf^2 - vi^2 = 2(g)(delta y) where the g would, if a numerical value is required, the appropriate sign would be input. So where a numerical value of g is not required, it would appear as thought I took g to be positive.
     
  7. Dec 22, 2013 #6
    Also, do you mind explaining the concept and derivation behind sqrt 2d/g?
     
  8. Dec 22, 2013 #7

    Curious3141

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    You edited after my post. Anyway, this correct. ##v_i## only differs in sign, the magnitude is the same. When you square a positive or negative number of the same magnitude, the result is the same.

    Another way to look at it (just for interest) is conservation of energy. When you throw the ball up, you impart a certain kinetic energy to it. As it goes upward, this energy gets progressively converted into gravitational potential energy that exists in the field between the ball and the Earth. At the max height, the ball is momentarily at rest, and all the energy is in the form of gravitational potential energy (zero kinetic energy). As it comes down again the reverse process occurs, and there is conversion of potential energy to kinetic energy. As the ball passes its former launching height travelling downward, it will have the exact same speed going down as it had going up (even though the direction is reversed). This is a consequence of the conservation of energy.

    A question to test your understanding: what "real world" condition can affect the trajectory so that this doesn't apply (and the ball will have a different speed going down than it did being thrown up)?

    Getting back to the kinematics treatment, when you calculate ##v_f## you need to take the square root of something. Remember there is a positive and a negative square root, and which you take depends on the physical interpretation. Which is correct here?
     
  9. Dec 22, 2013 #8

    Curious3141

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    Do you mean the kinematic equation ##v_f^2 -v_i^2 = 2as##? Better to clarify first.
     
  10. Dec 22, 2013 #9
    ±9.09ms^-1 but in ascribing downwards to imply a negative direction and upwards, positive, vi = -9.09ms^-1

    Are you familiar with solving this problem using the concept of limit?
     
  11. Dec 22, 2013 #10

    Curious3141

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    Should ##v_f## be positive? See my previous post.

    It's OK, whether you want to assign ##g = -9.81ms^{-2}## all the time, or you want to have the convention where ##g## just represents the numerical value, and you put the sign accordingly. Whatever it is, consistency is the key.
     
  12. Dec 22, 2013 #11
    Yes, I'm aware of this concept. Graphically, it takes on a sin graph, yes?
     
  13. Dec 22, 2013 #12
    It should be negative since the vf prior to hitting the ground is traveling in a downwards direction.
     
  14. Dec 22, 2013 #13

    Curious3141

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    What, exactly, follows a sine graph? Not everything that is periodic is a sine relationship. Some physical systems are periodic in a sinusoidal sense, like an oscillating idealised, friction-free, drag-free spring-mass system or (approximately) a pendulum exhibiting planar oscillations of small angular amplitude. But this system is not periodic, it just "retraces its path" the one time. And I can't see any relationship to the sine function.

    Not trying to knock you down, but one needs to be rigorous in one's statements.

    And what do you mean by solving with a limit concept? Not familiar with this for kinematics problems.
     
  15. Dec 22, 2013 #14
    Even if assuming acceleration due to gravity is 0ms^-1 and there exists no resistance of any sort?
    Wouldn't the object then reach a height, Δy, then back to where the ground, and repeats an infinite loop?

    What happens as the limit,t, tends towards the quadratic root t = 1.6......
     
  16. Dec 22, 2013 #15

    Curious3141

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    If acceleration was 0 (no gravitational influence), and there was no drag or any other force acting on it, the object would either remain at rest (if not thrown) or travel at a constant velocity after the initial throw. Same speed, same direction, forever. No parabolic path. It would never even reach the ground.

    And whether there is gravity or not, why would the object ever go in an infinite loop? With gravity, the object would hit the ground and stay there. Unless it had elastic properties.

    Not clear at all, sorry.
     
  17. Dec 22, 2013 #16
    Thanks. Informative. I missed one of your question earlier. Frictional force would result in the ball returning to a lower height than the max height achieved when it was thrown at the outset.
     
  18. Dec 22, 2013 #17

    Curious3141

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    Not really. Friction with the air (generally called drag) would result in the ball having a lower speed on its downward trajectory compared to its upward trajectory at the same height (horizontal level). Not sure why you think it will return to a lower height - the ball always hits a maximum height, drops to the ground and stays put there. Unless it bounces (the elastic property I mentioned earlier) - only then does it make sense to talk about returning to another height.
     
    Last edited: Dec 22, 2013
  19. Dec 23, 2013 #18
    I would really appreciate if you could shed some light on this:

    In part (a): vi = 7.3ms^-1 while delta y = -1.5m
    Why is delta y= -1.5m? Is there a rational for not measuring the displacement from ground to max height as opposed to max height to ground?
     
  20. Dec 23, 2013 #19
    part (c) When would the second ball hit the floor:

    Capture.JPG

    Why am I getting a negative t?

    The displacement has to be negative because yf < yi and indeed true when yf = 0m and yi = -1.5m above ground as the ball is falling towards the ground and we denote downwards as negative
     
  21. Dec 23, 2013 #20

    Curious3141

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    It's not max height (at least in the case of the ball being thrown upward), it's just the height of the hand above the floor.

    No particular rationale, you get to choose your convention, as long as you're consistent. If you want ##\Delta{y}## to be +1.5m, then your convention is that "up" is negative and "down" is positive. In which case, your acceleration will be +9.81 m/s^2 and your initial upward throw (for part a) will be with a velocity of -7.3 m/s.
     
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