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Motion Under Gravity

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data
    A ball is dropped from the top of a tower. If it takes 5 seconds to hit the ground, find the height of the tower and with what velocity will it strike the ground.

    Given Data:
    Initial Velocity = Vi = 0
    Acceleration = a = -g = 9.8 ms^-2

    2. Relevant equations
    The above question is my from my school textbook. I have the solution but I don't understand it. I want to know why the the acceleration (g in our case) is negative (minus).

    In my book it says that the acceleration is positive because the direction of initial motion is downward. When an object is speeding up, its acceleration is in the same direction as the velocity. All is fine till here. But if 'g' is positive then why is there a - in the data?

    To find the height of the tower I'll use the this equation

    S = Vi * t + 1/2 * g * t^2

    But in my book they have used above equation and solved it like this

    S = Vi * t - 1/2 * g * t^2
    S = 0 * 5 - 1/2 * -9.8 * 5^2
    S = 122.5 meters

    Why is there a - instead of +? and why is the g (why is 9.8 negative)?

    Finding the final velocity of the ball is easily done using this equation

    Vf = Vi + gt

    But here we use positive g i.e. 9.8 not -9.8

    Can you please explain?

    3. The attempt at a solution
     
  2. jcsd
  3. Apr 7, 2010 #2
    To find the height of the tower use equation:

    S = u.t + (1/2) a t^2

    and to find velocity use v = u + a.t
     
  4. Apr 7, 2010 #3

    Doc Al

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    Staff: Mentor

    It's better to start with:

    S = Si + Vi * t + 1/2 * a * t^2

    Usually, g stands for the magnitude of the acceleration due to gravity. So g = 9.8 m/s^2 (no minus sign).

    Gravity acts downward. Whether you call the acceleration (a) positive or negative depends on your sign convention. If up is positive, then a = -g. If down is positive, then a = g.
     
  5. Apr 7, 2010 #4
    Or for your question take this equations as:

    H = g. t^2

    and v = g. t
     
  6. Apr 7, 2010 #5
    Sorry, I answered you without looking down for your real problem.

    Sign convention actually make me go crazy. I never care for sign convention and make my own reasonable idea as a sign convention.

    Let consider your problem. In your problem you have dropped a stone from the top of the building. So you can easily find that the velocity of the stone is increasing with the time. So take g as positive.

    Now suppose You throw some stone upward. You may have noticed that the stone velocity decreases with time in this case you take g as negative.
     
  7. Apr 7, 2010 #6
    Sorry, I answered you without looking down for your real problem.

    Sign convention actually make me go crazy. I never care for sign convention and make my own reasonable idea as a sign convention.

    Let consider your problem. In your problem you have dropped a stone from the top of the building. So you can easily find that the velocity of the stone is increasing with the time. So take g as positive.

    Now suppose You throw some stone upward. You may have noticed that the stone velocity decreases with time in this case you take g as negative.

    Edit: Remember whenever we take "g" or "a" as positive we mean that body is accelerating (means velocity is increasing with time) and when "g" or "a" as negative we mean that body is retarding (means velocity is decreasing with time).
     
    Last edited: Apr 7, 2010
  8. Apr 7, 2010 #7

    Doc Al

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    Staff: Mentor

    Realize that sign conventions are completely arbitrary. You should be able to solve problems like these using either sign convention. If not, there's a gap in your understanding that should be fixed. Solve the problem both ways--prove to yourself that you get the same answer.

    That said, sometimes choosing a particular sign convention makes the arithmetic easier.
     
  9. Apr 7, 2010 #8
    yeah, try to solve using sign convention but don't make a habit to recall that sign convention every time you solve the problem. If you'll practice more and more problem, you'll find that sign convention become as easy as writing your name. (at least in one dimensional motion question).
     
  10. Apr 7, 2010 #9

    Doc Al

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    Staff: Mentor

    Careful here. In physics, a body is accelerating if its velocity is changing. Whether the acceleration is positive or negative depends on your sign convention--there's no implication that a positive acceleration means that speed is increasing or that a negative acceleration means speed is decreasing.

    Toss a ball straight up. It slows down until it reaches its maximum height, then begins speeding up as it falls. Yet the acceleration is always downward.
     
  11. Apr 7, 2010 #10
    Everything is vector here (except height). So it is difficult to explain. Acceleration due to gravity is always downward and no one can argue over it. But here it is not the case where the body returns back. In his case the body either goes up or down and so I've said that.

    Edit: And in the type of question You have given. I think it again starts a new section for the question because it return back to its initial position at the same speed at which it was thrown upward. only the direction is reversed and so he has a chance to shift the way he uses his sign convention.

    I've said that to give him some idea to trigger his practice. With practice he will know everything
     
    Last edited: Apr 7, 2010
  12. Apr 7, 2010 #11
    Here is a question I've found which should clear your problem of sign convention.

    A stone is thrown vertically upward with an initial velocity of 20 m/s. Find the time taken by the stone to return to the initial position.

    In this case, S = 0 (remember S is displacement not the distance traveled)

    a = - g (take it as 10 m/s^2 , actual value is 9.8 m/s^2)

    Using equation S = u.t + (1/2) a. t^2

    0 = 20 . t - 5 t^2

    0 = t (20 - 5. t)

    so t = 0 or t = 4 sec

    t = 0 for initially when the stone was not thrown up in the air and t = 4 for when the stone returns back to the initial position.

    Other method:

    Since it is going upward with an acceleration of -10 m/s^2 every second so after 1 second its velocity will be 10 m/s and then after 1 sec it will be 0 m/s and then it will start to fall. Now change the sign of g (as I've said in some of my earlier posts) so after 1 second velocity will be 10 m/s and after next second velocity will be 20 m/s.

    You can make it very clear from here that time taken to go up is same as time taken to return and also body returns with the same velocity with which it was thrown upward.

    Now I think it is very clear to you and You are not confused with the two different explanation given by me and Doc Al. Sign conventions are universally accepted but I thought you are confused with sign conventions as I was in my time so I've given you the second method to start practice for problems.
     
  13. Apr 7, 2010 #12
    Thank you all for clearing my confusions.
     
  14. Apr 8, 2010 #13
    I need help again.

    A ball is thrown vertically upward with a velocity of 98 m/s. How long does it take to reach its highest point?

    Time 't' can be obtained by using the equation

    Vf = Vi + a * t

    Vi = 98 m/s
    Since at highest point velocity is zero, therefore, Vf = 0
    If I take acceleration as negative i.e. a = -g = -9.8 m/s^2

    then we get,

    0 = 98 + (-9.8) * t
    -98 / -9.8 = t
    t = 10

    but if I take acceleration as positive i.e. a = g = 9.8 m/s^2

    0 = 98 + (9.8) * t
    -98 / 9.8 = t
    t = -10
    Okay, here the velocity decreases so we take acceleration as negative i.e. -9.8 m/s^2 and that makes sense. What in case we want to calculate the total time. For instance, a question of type to calculate how long does the ball remain in air? The velocity of the ball first decreases and then increases (falling down). We can calculate it as

    S = Vi * t * 1/2 * a * t^2

    Here do I take acceleration positive or negative?

    Now in this case which one is correct? I'm still confused :)
     
    Last edited: Apr 8, 2010
  15. Apr 8, 2010 #14

    Doc Al

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    Staff: Mentor

    This is fine. Here you take the sign convention that up is positive and down is negative. That means the initial velocity is positive and the acceleration is negative.

    You can't just make the acceleration positive, you have to use the complete sign convention: Down is positive and up is negative. That means the initial velocity is negative.


    Just pick a convention and stick to it throughout the problem. Using up as postive: The initial velocity will be positive and the acceleration will be negative.
     
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