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Projectile Motion

  1. Mar 18, 2004 #1
    For my mechanics, I am trying to model the flight of an objected released from an angled slide, which slides (not rolls) down a distance L (constant), launches at a hight h (constant) above ground and travels a distance X, untill it hits the ground. The slide's angle A is adjusted.

    My problem is I cannot seem to come up woith a simple formula, the only way I find to solve this using SUVAT equationsd as i have been told to ,ends up in a hefty quadratic equation, and several people have noted that there are simpler ways to model this? Any links to sites containing similar problems, or Ideas on how to approach this, and what equations to use would be greatly appriciated!

    http://www.skylinecomputers.co.uk/misc/Slide.JPG [Broken]

    A quick diagram I knocked up...
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Mar 18, 2004 #2


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    At the point of exit from the tube the problem becomes a falling body problem, this can be seperated into x and y components of motion. The first step is to define your coordinate system,let the origin be the ground level below the lower end of the tube, let y be vertical and increasing up, let x be horizontal and increasing away from the end of the tube. The end of the tube is at y=h.

    For the y or vertial motion you have:
    [tex] \frac {d^2y} {dt^2} = -g [/tex]
    With initial condions
    [tex] \frac {dy} {dt} \mid_0 = v_y [/tex]
    [tex] y(0) = h [/tex]

    for the x (horizontal motion) you have:
    [tex] \frac {d^2x} {dt^2} = 0 [/tex]
    With initial conditions:
    [tex]\frac {dy} {dt} \mid_0 = v_x [/tex]

    The equation for the y motion is simply a statement that the object falls with constant acceleration. The equation for x motion is a statement of no acceleration or constant velocity.

    The solution for the y motion is:
    [tex] y(t) = -\frac 1 2 g t^2 + v_yt + h[/tex]
    The solution for the x motion is :
    [tex] x(t) = v_x t [/tex]

    To get the initial velocity use conservation of energy,

    [tex] mgd = \frac 1 2 m v^2 [/tex]
    [tex] v = \sqrt {2gd}[/tex]
    d is the total drop of the tube.
    Using trig we get
    [tex] v_y = v \sin( \theta)[/tex]
    [tex] v_x = v \cos( \theta)[/tex]
    This is much more then I generally provide for this type of problem, but it is not clear what level you are working at (high school or college) I hope you can complete the problem by putting all the pieces together.
    Last edited: Mar 18, 2004
  4. Mar 18, 2004 #3


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    As long as the object is on the slide (and assuming no friction) the acceleration along the slide is -g sin A. The speed along the slide is -g sin(A)t and the distance traveled -g sin(A)t<sup>2</sup>/2. You could now calculate the time, t, required to travel the length of the slide, l, and use that to find the speed at the end of the slide. A simpler way is to use conservation of energy. Initially, the object has height l sin(A) and so potential energy, relative to the bottom of the slide, of lmg sin(A). At the bottom of the slide, all that energy is converted into kinetic energy: (1/2)m v<sup>2</sup>= lmg sin(A) so v<sup>2</sup>= 2lg sin(A). and v= &radic;(2lg sin(A)). Call that v<sub>0</sub>. After leaving the slide, the object has acceleration downward of -g and no horizontal acceleration. It velocity vector at any time t will be
    (v<sub>0</sub>cos(A), -gt+ v<sub>0</sub>sin(A)) and position vector (taking the bottom of the slide to be (0,h)) of (v<sub>0</sub>sin(A)t,-(g/2)t<sup>2</sup>+ v<sub>0</sub>sin(A)t+ h).

    The object "hits the sand" when -(g/2)t<sup>2</sup>+ v<sub>0</sub>sin(A)t+ h= 0. Yes, that's a quadratic equation but it can be solved using the quadratic formula. Once that has been done substitute into v<sub>0</sub>sin(A)t to find d.
  5. Mar 19, 2004 #4
    Thanks for the info, its for my a-level work... will go throught it ll when Iget time this evening and see if it helps... cheers.
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