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Projectile on an incline

  1. Mar 8, 2015 #1
    1. The problem statement, all variables and given/known data

    A projectile is thrown from a point A on an inclined plane field of unknown slope α, and hits the field on a point B. We know the angle between the initial velocity and the field: θ. We know the magnitude of the velocity: V. We know the the acceleration of gravity: g. We know the slant range of the projectile (distance between A and B): R. We have no friction.

    Calculate α in terms of θ,V,g, and R.

    I worked on this problem and I always get a complex 4th degree polynomial equation or a trigonometric equation that I can't figure out how to solve.

    2. Relevant equations
    V sin(alpha + theta)t = R sin(alpha)
    V cos (alpha+theta)t + (1/2)gt^2 = R cos(alpha)

    3. The attempt at a solution
    Consider alpha the angle between the gravity vector and the field. Consider a 2D reference frame (X,Y) that contains the trajectory and let Y be vertical and opposite to g and X be horizontal and such that the trajectory is in X > 0. if we decompose the equation of motion along X we get: V sin(alpha + theta)t = R sin(alpha). along Y we get: V cos (alpha+theta)t + (1/2)gt^2 = R cos(alpha). if we divide the second equation by the first we get a trigonometric equation with cot(alpha+theta), sin(alpha+theta), and cot(alpha). I couldn't go any further from here
     
  2. jcsd
  3. Mar 9, 2015 #2

    BvU

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    Hi jal, welcome (a bit belated) to PF :smile: !

    You have made an unconventional choice for ##\alpha## ! Usually we say ##\alpha = 0## means horizontal. For you it's straight down. Never mind.

    ##\theta=0## means parallel with the plane and I can't find a flaw in your story. Dividing 2nd by 1st still leaves a t which you eliminate with the first eqn. Leaving not only the ones you mention, but another ##\sin\alpha## as well.

    It is a single equation with a single unknown. I don't see a reasonable straightforward way towards an expression like ##\alpha = ...## (with no ##\alpha## on the RHS). I would solve it numerically.


    And I do wonder if just this single equation (with ##\alpha## on both sides) isn't enough answer for the composer of the exercise.

    Funny how such a simple situation can be made so complicated. After all, it's just a parabola passing to a suitable origin (point A) with an equation like y = P x ( x - Q) . Two degrees of freedom, three givens, so room for pinning down ##\alpha##.

    ---------------------------------------------------------------------------

    Now we follow Tuesdays advice by @A.T. from post #2 in this thread (which is a duplicate of the current thread -- PF frowns on that ?:) ! ) and decouple ##\theta## and ##\alpha## at the (small) cost of having uniformly accelerated motion in two directions instead of one:

    We say (going back to the more conventional ##\alpha = 0## and ##\theta = 0## when horizontal and I also like g = +9.81 m/s2, sorry) $$
    x = v_0\cos\theta\; t - {\tfrac 1 2} g \sin\alpha\; t^2 \\
    y = v_0\sin\theta\; t - {\tfrac 1 2} g \cos\alpha\; t^2 \\
    $$and know that ##y=0## at ##t=0## and at $$
    t={2v_o\sin\theta\over g \cos\alpha}
    $$You get somewhat cleaner equations, easier to type in :smile: when solving with e.g. excel :

    Parabola.jpg

    But I don't get much further than an equation like ##P\sin\alpha - Q\cos\alpha = C \cos^2\alpha## (with P, Q and C known) which for me is too difficult :smile:
     
    Last edited: Mar 9, 2015
  4. Mar 9, 2015 #3
    Thank you BvU for your reply!
    Not a complete solution but very useful :)
    I'll keep trying based on what u propose and see if I can get any further...
     
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