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Math Help - Projectile Motion

  1. #1
    Jul 2009

    Projectile Motion

    Here is a problem which i encountered, and have difficulties in terms of trying to solve.

    Question: A tall building stands on level ground. The nozzle of a water sprinkler is positioned point P on the ground at a distance d from a wall of the building. Water sprays from nozzle with speed V and the nozzle can be pointed in any direction from P.

    a) If V > sqrt (gd) prove that the water can reach the wall above ground level.

    My attempt:
    In this question, this is what i tried using standard vertical and horizontal motion, where t is time parameter, which i assumed was the approach as this is the current topic i am studying.

    t = d/(Vcosa) where a is varying angle.

    This is subbed into y = 1/2 gt^2 - Vtsin a

    Ultimately Ive obtained

    gd tan^2 a - 2V^2 tana + gd = y

    Here by using quadratic formula, for y=0 to exist, V > sqrt (gd)
    However, this solves for tan a, and i just cannot make the connection between V>sqrt gt and y is able to reach wall above ground level.

    I assume this solution would depend upon inequality of
    gd tan^2 a - 2V^2 tana + gd > 0,

    But this still solves for tan a. Is my method flawed, as I am thinking there may be another way without using tan a?
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  2. #2
    MHF Contributor
    skeeter's Avatar
    Jun 2008
    North Texas
    since water from the hose is projected from ground level, the max horizontal range is achieved when the launch angle = 45 degrees, so \tan(a) = \tan(45) = 1<br />

    also, y = v\sin(a) \cdot t - \frac{1}{2}gt^2 (you had the signs reversed)

    sub in \frac{d}{v\cos(a)} for t ...

    y = d\tan(a) - \frac{gd^2}{2v^2}\tan^2(a) - \frac{gd^2}{2v^2}

    since tan(a) = tan(45) = 1 ...

    y = d - \frac{gd^2}{v^2}

    for v = \sqrt{gd} ...

    y = d - \frac{gd^2}{gd} = d - d = 0

    for v > \sqrt{gd} , d > \frac{gd^2}{v^2} which implies d > 0
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