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A body thrown upwards from a stand of height

  1. Nov 27, 2011 #1
    I want to check whether my solution is right or not... beside I have made some correction to the problem so that it will be consistent ( I want to check also this correction).:smile:

    1. The problem statement, all variables and given/known data

    A body is thrown vertically upwards from a stand of height 30 meters above the ground. If the height h(t) above the ground after t seconds is given by : h(t) =20t-5t^2 +25 meters . Find:



    (a) The initial velocity and acceleration and also the velocity and acceleration after t seconds.

    (b) The highest point reached by the body.

    (c) When the body strikes the ground and its velocity then.

    (d) Suppose that instead of being thrown from a stand the body was thrown vertically from the ground with the same acceleration and initial velocity . When will the body strike the ground and find its velocity then.



    ******************************
    This problem is incorrectly formulated:- (I am not sure about that)

    I have copied this problem as it is,, and I think that there is a mistake in the problem formulation

    Since the stand's height is 30 meters above the ground, the equation should be formulated like this
    h(t) =20t-5t^2 +30 meters.

    OR instead (that what I will do) correct the height of the stand from 30 to 25 so that the corrected version will be:

    A body is thrown vertically upwards from a stand of height 25 meters above the ground. If the height h(t) above the ground after t seconds is given by : h(t) =20t-5t^2 +25 meters.

    ******************************
    2. Relevant equations

    3. The attempt at a solution

    (a) The initial velocity and acceleration and also the velocity and acceleration after t seconds.

    the velocity:

    v(t) = (dh/dt) = 20-10t m/s

    initial velocity is when t=0

    v(0) = 20 m/s

    the acceleration:

    a(t) = (dv/dt) = -10 m/s^2

    so a=-10 m/s^2

    ....................................

    (b) The highest point reached by the body.

    the highest point is when v(t) = 20-20t = 0

    20-10t=0

    t = 2 seconds

    the highest point can be given using h(t) =20t-5t^2 +25

    h(2) =20(2)-5(2)^2 +25 = 45 m

    ......................................


    (c) When the body strikes the ground and its velocity then.

    the body strikes the ground when h(t) =20t-5t^2 +25 = 0

    20t-5t^2 +25 = 0

    using the quadratic formula

    t = -1 seconds and t = 5 seconds

    **************************
    I want to know what each result represents, and what is the meaning of the minus sign in (t=-1)
    **************************
    the velocity can be given

    v(t) = 20-10t m/s

    v(-1) = 30 m/s

    v(5) = -30 m/s

    **********************
    I want to know what each result represents, and what is the meaning of the minus sign in v(5) = -30 m/s
    **********************
    .............................

    (d) Suppose that instead of being thrown from a stand the body was thrown vertically from the ground with the same acceleration and initial velocity . When will the body strike the ground and find its velocity then.

    Here I changes the equation of height to be
    h(t) =20t-5t^2

    the equation of velocity (not changed)
    v(t) = 20-10t m/s


    the body will strike the ground when h(t) =20t-5t^2 = 0

    t(20-5t) = 0

    t=0 seconds and t = 4 seconds

    **************************
    I want to know what each result represents.
    **************************

    the velocity can be given v(t) = 20-10t m/s

    v(0) = 20 m/s

    v(4) = -20 m/s

    **************************
    I want to know what each result represents. and what is the meaning of the minus sign in v(4) = -20 m/s
    **************************
     
    Last edited: Nov 27, 2011
  2. jcsd
  3. Nov 27, 2011 #2
    Hi there.
    Unless there's some convoluted trick nestled in the query, such as having the object thrown with some initial displacement, while already on the stand, I think you're positively right in the corrections you provided, though +25 seems more accurate, since it produces very neat and tidy results(in the quadratic equation).
    Next, it seems to me that you've solved everything accurately; Note an easy form of validation: your acceleration a=-10, hence -g, which what you would expect for a body under the 'auspices' of earth's gravity;
    As for the two roots, try sketching the h(t) parabola, taking t, the horizontal axis, from some negative value to a positive one, and observing the result, Keep in mind, that when it comes to time, it is our physical convention that for a physical process, some event starts at 0. However, there "was"(perhaps) something before, and if we allow it to continue(and conditions permit), there will be some "after"(literally, after the object has hit the ground in this case).
    The negative results when plugging in the velocity are evident, someone has already chosen the sign convention for you in this problem; That is, a-the acceleration is already taken to be negative, so when the object returns to the ground, it has a velocity matching the -y direction.
    Hope that clears things up,
    Daniel
     
  4. Nov 27, 2011 #3

    PeterO

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    Homework Helper

    All answers included above.
     
  5. Nov 29, 2011 #4
    Thank you very much : "Daniel" "PeterO"

    It is clear now. :smile:
     
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