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    #1
    a. If a flea can jumps straight up to a height of 0.440m.
    what is the initial speed as it leaves the ground.
    b. For how much time is it in the air?

    ok I am sure we use this equation

    $$x=x_0+v_o^t+\frac{1}{2} at^2$$

    I watched another YT on this but it got to much complication
    better just to come here

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    #2
    Apart from YouTube, do you also have a textbook? What does that say?
    (I am asking because I am rarely able to understand something from a video that I did not understand previously already.)

    I think you mean $v_0t$ in the equation for $x$?

    In a drawing, let the vertical axis be the $x$ axis, with upward = positive. The initial $x$-position $x_0 = 0$.
    What do you choose for the acceleration $a$? What is the sign of $a$?

    Now note that at the maximal $x$-position the velocity equals zero.
    How do you obtain an expression for the velocity from the expression you have given?

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    #3 Thread Author
    I got that from the YT vid

    the text is University Physics
    but equation the is also form 2-12 on their eq list

    I not in a class just doing this on my own.
    Last edited by karush; December 19th, 2017 at 23:15.

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    #4
    (a) $v_f^2 = v_0^2 -2g \Delta y$

    At the top of its trajectory, the flea’s velocity is zero and $\Delta y = h_{max}$ ...

    $v_0 = \sqrt{2g h_{max}}$

    (b) calculate the time it takes for the flea to reach the top of its trajectory, $h_{max}$, and double it ...

    $v_f = v_0 - gt \implies t = \dfrac{v_0}{g}$

    total time, $T=\dfrac{2v_0}{g} = \dfrac{2\sqrt{2gh_{max}}}{g}$

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    #5 Thread Author
    ​
    Quote Originally Posted by skeeter View Post
    (a) $v_f^2 = v_0^2 -2g \Delta y$

    At the top of its trajectory, the flea’s velocity is zero and $\Delta y = h_{max}$ ...

    $v_0 = \sqrt{2g h_{max}}$

    (b) calculate the time it takes for the flea to reach the top of its trajectory, $h_{max}$, and double it ...

    $v_f = v_0 - gt \implies t = \dfrac{v_0}{g}$

    total time, $T=\dfrac{2v_0}{g} = \dfrac{2\sqrt{2gh_{max}}}{g}$
    ok the text book answers to this is

    a. 2.94 m/s
    b. 0.599 s

    I'll see if i can plug in to get em
    $\displaystyle 0.599s=\frac{2\sqrt{2gh_{max}}}{g} $
    if $g=-9.8$ and $h_{max}=.44$
    Last edited by karush; December 20th, 2017 at 17:14.

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    #6
    Quote Originally Posted by karush View Post
    a. If a flea can jumps straight up to a height of 0.440m.
    what is the initial speed as it leaves the ground.
    b. For how much time is it in the air?

    ok I am sure we use this equation

    $$x=x_0+v_ot+\frac{1}{2} at^2$$

    I watched another YT on this but it got to much complication
    better just to come here
    Well, since the "up to a height of 0.44m" is the height above the starting height, you can take $x_0= 0$. Then $x= v_0t+ \frac{1/2}at^2= \frac{1}{2}a(t^2+ \frac{2v_0}{a}t)$ is a quadratic and we can "complete the square": $x= \frac{a}{2}\left(t^2+ \frac{2v_0}{a}t+ \frac{v_0^2}{a^2}- \frac{v_0^2}{a^2}\right)= \frac{a}{2}\left(x+ \frac{v_0}{a}\right)^2- \frac{v_0^2}{2a^2}$. "a" here is negative so that this is an "upside down" parabola. When $t= -\frac{v_0}{a}$ it takes its maximum value $-\frac{v_0^2}{2a}$.

    For (a), assuming this is on the surface of the earth, set a= -9.81 and solve $\frac{v_0^2}{19.62}= 0.440$ for $v_0$.

    For (b), since a parabola is symmetric about its midline, take t equal to twice the "t" above, $-\frac{2v_0}{a}$.
    Last edited by HallsofIvy; December 23rd, 2017 at 16:08.

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    #7 Thread Author
    Quote Originally Posted by HallsofIvy View Post
    Well, since the "up to a height of 0.44m" is the height above the starting height, you can take $x_0= 0$. Then $x= v_0t+ \frac{1/2}at^2= \frac{1}{2}a(t^2+ \frac{2v_0}{a}t)$ is a quadratic and we can "complete the square": $x= \frac{a}{2}\left(t^2+ \frac{2v_0}{a}t+ \frac{v_0^2}{a^2}- \frac{v_0^2}{a^2}\right)= \frac{a}{2}\left(x+ \frac{v_0}{a}\right)^2- \frac{v_0^2}{2a^2}$. "a" here is negative so that this is an "upside down" parabola. When $t= -\frac{v_0}{a}$ it takes its maximum value $-\frac{v_0^2}{2a}$.

    For (a), assuming this is on the surface of the earth, set a= -9.81 and solve $\frac{v_0^2}{19.62}= 0.440$ for $v_0$.

    For (b), since a parabola is symmetric about its midline, take t equal to twice the "t" above, $-\frac{2v_0}{a}$.
    2(2.94)/9.81=.599

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    #8
    Quote Originally Posted by karush View Post
    a. If a flea can jumps straight up to a height of 0.440m.
    what is the initial speed as it leaves the ground.
    b. For how much time is it in the air?

    ok I am sure we use this equation

    $$x=x_0+v_o^t+\frac{1}{2} at^2$$

    I watched another YT on this but it got to much complication
    better just to come here
    For part a), we could use conservation of energy...initially the flea has kinetic energy, and then when it reaches the apex of its jump, it has potential energy:

    $ \displaystyle \frac{1}{2}mv_0^2=mgh$

    Solving for $v_0$, we obtain:

    $ \displaystyle v_0=\sqrt{2gh}$

    For part b), we could write:

    $ \displaystyle h(t)=\frac{t}{2}\left(2v_0-gt\right)$

    We are interested in the non-zero root, hence:

    $ \displaystyle t=\frac{2v_0}{g}=\frac{2\sqrt{2gh}}{g}=2\sqrt{\frac{2h}{g}}$

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