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Thread: 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 $\displaystyle h$ positive
    upward and the origin as the point from which it is threown we have:

    $\displaystyle h''=-g$

    so:

    $\displaystyle h'=-gt+v_0$

    and :

    $\displaystyle
    h=-gt^2/2+v_0t
    $

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

    $\displaystyle
    -50 = -18 g +6 v_0
    $

    so:

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

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

    $\displaystyle t=v_0/g$

    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? . $\displaystyle y \;=\;h_o + v_ot - 4.9t^2$
    . . where $\displaystyle h_o$ = initial height, $\displaystyle v_o$ = initial velocity,
    . . and $\displaystyle y$ = height of object at time $\displaystyle t$.

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


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

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


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

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

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

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

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

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

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

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

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

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    thank you!
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