Results 1 to 2 of 2

Math Help - Projectile Motion

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    132

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. projectile motion
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: June 29th 2010, 06:54 AM
  2. Projectile Motion
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 22nd 2010, 03:12 AM
  3. Projectile Motion #1
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: September 27th 2009, 01:52 PM
  4. Projectile motion (1)
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: August 18th 2009, 07:08 AM
  5. Replies: 1
    Last Post: February 19th 2009, 02:30 PM

Search Tags


/mathhelpforum @mathhelpforum