As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your wind

Question

As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is Lw. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g. Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward). If the bottom of your window is a height hb above the ground, what is the velocity vground of the pot as it hits the ground? You may introduce the new variable vb, the speed at the bottom of the window, defined by
vb = Lwt + gt2.

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bonexptip 1 year 2021-08-29T21:06:26+00:00 1 Answers 21 views 0

Answers ( )

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    2021-08-29T21:08:22+00:00

    Answer:

    [tex]\mathbf{v_{ground} = \sqrt{{v^2+2ghb}}}[/tex]

    Explanation:

    From the information given:

    The avg. velocity post the window is;

    [tex]v_{avg} = \dfrac{L_w}{t}[/tex]

    [tex]v_b[/tex] = velocity located at the top of the window

    [tex]v_b[/tex] = velocity situated at the bottom of the window  

    Using the equation of kinematics:

    [tex]v_b = v_t + gt[/tex]

    Hence,

    [tex]v_t = v_b – gt[/tex]

    To determine the average velocity as follows:

    [tex]v_{avg} = \dfrac{1}{2} (v_t + v_b)\dfrac{L_w}{t}= \dfrac{1}{2}(v_b – gt +v_b) \\ \\\dfrac{L_w}{t} = v_b – \dfrac{1}{2}gt \\ \\ v_b = \dfrac{L_w}{t }+ \dfrac{1}{2} gt\\ \\ = \dfrac{1}{t} \Bigg(L_w + \dfrac{1}{2}gt^2 \Bigg) \\ \\[/tex]

    where;

    [tex]v_b[/tex] = velocity gained when fallen through the height h.

    Similarly, using the equation of kinematics, we have;

    [tex]v_b^2 = 2gh \\ \\h = \dfrac{v_b^2}{2g}[/tex]

    [tex]\implies \dfrac{(L_w + \dfrac{1}{2} gt^2_^2}{2gt^2}[/tex]

    Thus, the velocity at the ground is;

    [tex]v^2_{grround} = v_b^2 + 2ghb[/tex]

    [tex]\mathbf{v_{ground} = \sqrt{{v^2+2ghb}}}[/tex]

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