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B Quick throwing and dropping ball Question

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  1. Sep 2, 2016 #1
    I am relatively new in understanding physics. I have a quick question that I think a lot of you will find easy to explain. Why is it that when a ball is dropped initial velocity is 0 but when it is thrown the initial velocity is not 0? Do they not both start at rest?
     
  2. jcsd
  3. Sep 2, 2016 #2
    First of all, welcome to the forum.
    If you just began a course in physics, such a thing will be addressed in a later lesson within the course, typically they teach you how things move before telling you why they move, first for historical reasons and second that it is easier to grasp than what one famous mathematician in England has came up with to describe motion, which also should be mentioned and talked about.
    In general, when you throw the ball you push it, giving it velocity. Essentially it was accelerated, but now in a free fall under the influence of one source of acceleration, which will be named later in the course, and you want to describe that very motion, you have to take the acceleration that was before the problem started into account by giving it the 'initial condition', a constant which states the beginning of the problem. For this one it would be the velocity it got before it was released in the throw.
    So while previously the ball WAS at rest as the implication "thrown" is there, for the purposes of describing and solving the problem it has not began at rest, it was given the velocity before the events which you are focusing on.
    'At rest' means without a velocity in physical problems. Many things used in physics sometimes mean different things in other places. You will get used to the terminology as you progress.
     
    Last edited: Sep 2, 2016
  4. Sep 2, 2016 #3
    Thank you for the quick response. I just have it in my mind that initial velocity is the velocity at time t = 0. I find it weird how 'the holding' of the ball is considered t = 0 but the throwing of the ball(after the hold) is considered t = 0. I guess it just takes time to learn how to interpret things in terms of physics. :D
     
  5. Sep 2, 2016 #4
    Oh and thank you for welcoming me to the forum. I see myself being on here a lot because I do like physics and would like to really understand it.
     
  6. Sep 2, 2016 #5
    You are welcome, wish you success in physics.
     
  7. Sep 2, 2016 #6

    sophiecentaur

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    I have to say 'It all depends', I'm afraid and there isn't a 'rule' to tell you what to do.
    In fact, you can choose to take the origin of your time to be anything - last Thursday night at 10 pm exactly - or the instant you start the throwing action or the instant the ball is actually released. Which you choose will depend on what you actually want to find out from the exercise. You will get the same answer if you calculate its position at any time relative to the origin you choose. It is normal to choose the initial situation to be as simple as possible - often, the instant the object is released but there are situations where it's necessary to include the speeding up phase, before the object is released - for instance, the distance that the arm is travelling could made a big difference to the final range achieved.
     
  8. Sep 2, 2016 #7

    CWatters

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    +1

    In many problems you aren't interested in how the projectile gets its initial velocity, you are only interested in what the initial velocity is. For example the distance that a cannon ball travels depends on the velocity with which it leaves the cannon. If you have that info you can ignore the details of how it gets to that velocity. In such cases its convenient to define the "initial velocity" as the velocity with which it leaves the cannon. If you had enough information you could pick another point in time to use as the initial velocity. It need not be zero, it could be the velocity with which the cannon ball left the factory where it was made although that would unnecessarily complicated the calculations. With practice you will get used to dividing up problems into suitable time intervals and learn what's relevant or not to solving the problem.

    Oh yes.. Welcome to the forum!
     
  9. Sep 2, 2016 #8
    If you guys can help me understand this, that would be very helpful. When you drop an object, say a ball from rest, the initial velocity is 0. What would it mean if the initial velocity wasn't 0? And why would that take longer to hit the ground?

    Numbers:
    ---initial velocity set to 0---
    Initial height 20.0m
    Initial Velocity 0 m/s
    Gravity 10.0m/s/s
    t = 2.00 s
    y = 0.00m
    v =-20.00m/s



    ---initial velocity not 0---
    Initial height 20.0m
    Initial Velocity 2.0m/s
    Gravity 10.0m/s/s
    t = 2.21 s
    y = 0.00m
    v =-20.10m/s


    The reason why I ask this question is because in lab we were doing a 1-Dimensional problem where we dropped a ruler and wanted to calculate the time it fell. I initially wanted to set the final velocity to 0(hit ground) and find the initial velocity with my givens(vf,a,change in y), but it turned out that by doing that I am changing the problem and therefore would get the wrong results.
     
    Last edited: Sep 2, 2016
  10. Sep 2, 2016 #9
    As sophiecentaur indicated, you have flexibility about the point you choose as the start of the problem. In a case like you describe, the starting point is dictated by the fact that the equations you are probably going to use probably assume that acceleration is constant (and equal to the acceleration due to gravity). That is true from the moment of release to the instant before reaching the ground. At the moment of release, the velocity is zero.

    Do you see that the ruler's acceleration changes when it touches the ground? You could take the final velocity to be 0, but that would require breaking the problem into two parts:
    a) from release to just reaching the ground (where the initial value is zero, the acceleration is due to gravity and you would have to calculate the final velocity), and
    b) from contacting the ground until it stops (so the initial velocity is the final velocity from the part a, the acceleration is not due to graviyy and the final velocity is zero.

    Note that you would need to know more information (like stopping time, stopping distance, stopping acceleration...) to work with the second part.
     
  11. Sep 3, 2016 #10

    CWatters

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    There are three cases...

    Object thrown upwards (eg initial velocity not zero).
    Object dropped (eg initial velocity zero).
    Object thrown downwards (eg initial velocity not zero).

    The first will take longer to hit the ground than the second as the object goes up before it starts down.
    The third will take less time to hit the ground than the second.

    Normally you start this type of problem by defining one direction as "positive". So if we define downwards as positive then the initial velocity in the first case is negative. It's important to pay attention to the sign of everything when you plug numbers into the equations.
     
  12. Sep 6, 2016 #11

    PeterO

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    Don't forget that time t=0 is an arbitrary time you have chosen, and then you are going to examine motion etc from then on. If you stand atop a freeway overpass, and decide that 10:00am is what you will define at time t=0, this does not mean that the cars on the freeway cannot move until 10:00 am arrives. It is just that at time t=0, some cars are already travelling at, perhaps, 25 metres per second. The broken-down truck parked in the emergency lane may have velocity of zero, however.
     
  13. Sep 6, 2016 #12

    PeterO

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    In your examples, I would prefer to see Acceleration given as -10.00., The initial height of 20, final height of 0 and final velocity of -20m/s all imply/express down as a negative direction. The acceleration due to gravity is certainly also down, so should be listed as -10.0 m/s/s.
     
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