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24 December, 07:19

a flower pot is theown out of a window with a horizontal velocity of 8 m/s. If the window is 1.5 m off the ground, how far from the window does it land?

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Answers (2)
  1. 24 December, 07:27
    0
    T = The time it takes for the flower pot to pass the top of my window.

    V = The velocity of the flower pot at the moment it is passing the top of my window.

    X = The height above the top of my window that the flower pot was dropped.

    h = Lw + X

    Lw = (1/2) * g*t^2 + V*t

    V*t = Lw - (1/2) * g*t^2

    V = Lw/t - (1/2) * g*t, On the other hand we know : V=gT.

    Therefore we will have: Tg = Lw/t - (1/2) * g*t

    T = Lw / (tg) - t/2

    Now substitute for T in the following equation: X = (1/2) * g*T^2

    X = (1/2) * g * (Lw / (tg) - t/2) ^2

    Now substitute for X in the very first equation I mentioned: h = Lw + X

    h = Lw + (1/2) * g * (Lw / (tg) - t/2) ^2

    In case you wanted the answer to be simplified, then:

    h = (Lw^2) / (2*g*t^2) + (g*t^2) / 8 + Lw/2
  2. 24 December, 07:48
    0
    3.26m

    Explanation:

    Using one of the equation of motion to get the distance of the pot from the window and the ground;

    v² = u²+2as where

    v is the final velocity = 8m/s

    u is the initial velocity = 0m/s

    a = + g = acceleration due to gravity (this acceleration is positive since the body is falling downwards)

    g = 9.81m/s

    s is the distance between the object and the window from which it dropped.

    Substituting this values to get the distance s we have;

    8² = 0²+2 (9.81) s

    64 = 19.62s

    s = 64/19.62

    S = 3.26m
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