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A question about Motion in One & Two Dimensions

  1. Jan 14, 2015 #1
    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

    A balloon going upwards with a velocity of 12 m/s, is at a height of 65m from the earth at any instant. Exactly at this instant a packet drops from it. How much time will the packet take in reaching the earth.
    Ans. as given in my school book is: 5sec.
    But my problem is that I am not able to understand how to solve it. It will be great if somebody would please help me out and provide me the procedure for solving it.
    Thankyou.
     
    Last edited by a moderator: Jan 14, 2015
  2. jcsd
  3. Jan 14, 2015 #2

    Stephen Tashi

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    What equations do you know that apply to an object moving with a constant acceleration?
     
  4. Jan 14, 2015 #3
    1. v=u+at
    2. s=ut+0.5at( t raised to the power 2)
    3. v(raised to the power 2)=u(raised to the power 2)+ 2as

    where v=final velocity
    u=initial velocity
    a= acceleration
    t= time
    s= distance/displacement/height
     
    Last edited: Jan 14, 2015
  5. Jan 14, 2015 #4

    Stephen Tashi

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    In the notation, I am familiar with:

    a: the constant acceleration
    v(t): velocity at time t
    v0: velocity at time t = 0
    d(t): distance at time t
    d0: distance at time t = 0

    1) v(t)= v0 + at
    2) d(t)= d0 + v0 t + 0.5 a t^2

    You must choose the event that defines the time t = 0. Let's say t = 0 at the instant the package is dropped.
    You must choose which direction is positive. Let's say that upward is positive and downward is negative.
    You must choose what distance to call the zero distance. Let's say that ground level is zero.

    What is a ?
    What is v0 ? (u in your notation).
    What is d0 ? ( s at time t = 0 in your notation)
     
  6. Jan 14, 2015 #5
    Sir, my problem is this only. I get horribly confused while taking a sign convention for equations of motion in numericals of projectile motion.
    Suppose if i choose a vector going upward to be positive and vector going downward to be negative.
    Then in this situation,if a package is dropped from the baloon...........as I have tried to do it...
    u= -12m/s ( bcoz u is directed downwards)
    s=65m
    I had used the equation:
    s=ut+ 0.5at(t raised to the power 2)
    Here a = g i.e 9.8m/s2
    But acc. to sign convention choosen by me what should be the sign for g???? I should take ' - 9.8 ' or ' +9.8 ' as the acceleration due to gravity????? Plz help me out sir.......If this confusion of mine gets cleared, I will easily solve all the other numericals of this chapter in my book.My book is full of such numericals and I seriously need someone to clear out this sign convention application's confusion......
     
  7. Jan 14, 2015 #6
    g is directed downwards. What does that tell you? (Hint: you said what it should be when you gave a previous value)
     
  8. Jan 14, 2015 #7

    Stephen Tashi

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    The box is moving upward with the balloon at t = 0., u = +12 m/sec

    Do you mean s= 65 m at t = 0 ? That's ok if we say the distance s = 0 is ground level.

    If you mean s to be the change in distance between the position of the package when it is released and when it hits the ground then S= -65 m since the change is a decrease in distance relative to the posiive upward direction.

    That equation will work if you say that s = 0 when t = 0. By that convention, the zero distance is the position of the balloon when the package is released. The final position is at s = -65.

    That equation will also work if you interpret s to be "the change in distance" from time 0 to time t instead of saying s is "the distance at time t".

    I prefer to say s = s0 + ut + 0.5 a t^2 where s(t) is "the distance at time t", s0 = 65 is the distance at time t = 0
    Then solve for the time when s = 0, which is where the package hits the ground.

    If you want to leave s0 out of the equation, then you are dealing with a change in distance . You'd solve for the time when there is a change in distance of -65 m.

    It's very important to keep in mind whether a quantity in an equation represents and "absolute" measurement (e.g. "distance") or a relative measurement (e.g. "change in distance" ). A measurement requires defining where the zero of the measurement is. A change in a measurement is a change relative to what the measurement is at some initial state (often picked to be the state when t = 0).

    Use a negative acceleration a = -9.8 m/sec^2
     
  9. Jul 13, 2017 #8
     
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