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Math Help - A problem about free fall

  1. #1
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    A problem about free fall

    An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point 50.0m below its starting point 6.00 seconds after it leaves the thrower's hand. Air resistance may be ignored.
    a.) what is the initial speed of the egg? (answer: 21.2 m/s)


    b.)How high does it rise above its starting point?(answer: 22.6m)

    im stuck with these two questions and i got 58m/s in the 1st part ...im reallyyy stuck
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point 50.0m below its starting point 6.00 seconds after it leaves the thrower's hand. Air resistance may be ignored.
    a.) what is the initial speed of the egg? (answer: 21.2 m/s)


    b.)How high does it rise above its starting point?(answer: 22.6m)

    im stuck with these two questions and i got 58m/s in the 1st part ...im reallyyy stuck
    You need the equation of motion which if we take h positive
    upward and the origin as the point from which it is threown we have:

    h''=-g

    so:

    h'=-gt+v_0

    and :

    <br />
h=-gt^2/2+v_0t<br />

    We are told that h=-50 at t=6 so:

    <br />
-50 = -18 g +6 v_0<br />

    so:

    v_0=\frac{18g-50}{6}

    It reaches its maximum height when h'=0 which is when:

    t=v_0/g

    RonL
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  3. #3
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    Hello, ^_^Engineer_Adam^_^!

    An egg is thrown nearly vertically upward from a point near the cornice of a tall building.
    It just misses the cornice on the way down and passes a point 50.0m below its starting point
    6.00 seconds after it leaves the thrower's hand.

    a) What is the initial speed of the egg? (answer: 21.2 m/s) ?

    b) How high does it rise above its starting point? (answer: 22.6m)

    Are you familiar with the free-fall formula? . y \;=\;h_o + v_ot - 4.9t^2
    . . where h_o = initial height, v_o = initial velocity,
    . . and y = height of object at time t.

    Let launch point be at h_o=0.
    . . Then we have: . y \:=\:v_ot - 4.9t^2


    (a) We are told that y = -50 when t = 6.
    . . So we have: . -50 \;=\;v_o(6) - 4.9(6^2)\quad\Rightarrow\quad v_o \;=\;\frac{126.4}{6} \;=\;21.0666...

    Therefore, the initial speed is: . v_o \;\approx\;\boxed{21.1\text{ m/s}}


    (b) To maximize height, equate the derivative to zero.

    We have: . y \;=\;21.1t - 4.9t^2
    Then: . y' \;=\; 21.1 - 9.8t \;=\;0\quad\Rightarrow\quad t \;=\;\frac{21.1}{9.8} \;\approx\;2.15 seconds

    Therefore: . y \;=\;\frac{316}{15}(2.15) - 4.9(2.15)^2 \;\approx\;\boxed{22.6\text{ feet}}

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

    If you aren't allowed to use Calculus, we can still solve it.

    The graph of: . y \;=\;21.1t - 4.9t^2 is a down-opening parabola.
    . . Its maximum height occurs at its vertex.

    The vertex is given by: . t \:=\:\frac{-b}{2a}

    We have: . a = -4.9,\;b = 21.1

    . . Hence: . t \;=\;\frac{-21.1}{2(-4.9)} \;\approx\;2.15 seconds . . . see?

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  4. #4
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    thank you!
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