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Math Help - [SOLVED] Initial height question in vector spaces

  1. #1
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    [SOLVED] Initial height question in vector spaces

    A stone is thrown from a rooftop at time t = 0. Its position at time t is given by r(t) = (10t,−5t, 6.4−4.9t^2). The origin is at the base of the building, which is on flat ground. Distance is measured in meters and time in seconds.
    (a) How high is the rooftop?

    I know that the initial velocity r'(t) = 0 at t = 0. But I am unsure on how to solve for the height of the building using parametric equations.
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  2. #2
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    Hello, unifieda!

    This is a strange one . . . an "eyeball" problem
    Just look at it . . .nothing fancy is needed!


    A stone is thrown from a rooftop at time t = 0.
    Its position at time t is given by: .  r(t) \:=\:(10t,\: -5t,\: 6.4-4.9t^2)
    The origin is at the base of the building, which is on flat ground.
    Distance is measured in meters and time in seconds.

    (a) How high is the rooftop?

    I know that the initial velocity r'(t) = 0 at t = 0. . This is not true!
    We are given three functions: . \begin{Bmatrix}x \;=\;10t \\ y \;=\;-5t \\ z \;=\;6.4-4.9t^2 \end{Bmatrix}


    I would guess the following:

    x is the east-west location of the stone.
    . . It is moving east at 10 m/sec. **

    y is the north-south location of the stone.
    . . It is moving south at 5 m/sec. **

    z is the height of the stone above the ground.


    At t = 0,\;z = 6.4
    The stone began at 6.4 m above the ground.

    . . Therefore, the height of the building is 6.4 meters.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    The stone is thrown horizontally from the building.
    Looking down, the scene looks like this:
    Code:
               10t
        A * - - - - - *
             *  θ     |
                *     | 5t
                   *  |
                      * B

    The stone is thrown from A in the direction of B.

    Its speed is: . \sqrt{10^2 + 5^2} \:=\:\sqrt{125} \:\approx\:11.2 m/sec.

    Its direction is: . \theta \:\approx\;26.7^o south of east.


    So, you see, the initial velocity was not zero.

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, unifieda!

    This is a strange one . . . an "eyeball" problem
    Just look at it . . .nothing fancy is needed!


    We are given three functions: . \begin{Bmatrix}x \;=\;10t \\ y \;=\;-5t \\ z \;=\;6.4-4.9t^2 \end{Bmatrix}


    I would guess the following:

    x is the east-west location of the stone.
    . . It is moving east at 10 m/sec. **

    y is the north-south location of the stone.
    . . It is moving south at 5 m/sec. **

    z is the height of the stone above the ground.


    At t = 0,\;z = 6.4
    The stone began at 6.4 m above the ground.

    . . Therefore, the height of the building is 6.4 meters.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    The stone is thrown horizontally from the building.
    Looking down, the scene looks like this:
    Code:
               10t
        A * - - - - - *
             *  θ     |
                *     | 5t
                   *  |
                      * B

    The stone is thrown from A in the direction of B.

    Its speed is: . \sqrt{10^2 + 5^2} \:=\:\sqrt{125} \:\approx\:11.2 m/sec.

    Its direction is: . \theta \:\approx\;26.7^o south of east.


    So, you see, the initial velocity was not zero.

    Thank you so much for the help. I cannot believe I did not see this.
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