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Free fall motion

  1. Sep 10, 2003 #1
    Consider a particle located at (x0, y0), having inital velocity v0, and the angle of inclination of the velocity vector is [the]. The particle is subjected to a vertical acceleration of -g. It's equation of motion is...

    x = x0 + v0*cos[the]*t
    y = y0 + v0*sin[the]*t - 0.5*g*t2

    If t>0, the particle will reach y = 0 when
    t = {v0*sin[the] + sqrt((v0*sin[the])2+2*g*y0)}/g

    (found using the abc formula)

    If we call the time t1, then at that time the x position of the particle is (call it x1)...

    x1 = x0 + v0*cos[the]*t1

    The question is: at initial height y0, what angle will make x1 maximum?

    I found it very hard to make the equation simple enough to be able to solve this problem... Can anyone help me?

    PS: in the case of y0 = 0, the problem will be easy to solve, with the answer 45 degrees.

    Thank you...
     
    Last edited: Sep 10, 2003
  2. jcsd
  3. Sep 11, 2003 #2

    HallsofIvy

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    Replacing t1 in x1= x0+ v0 cos[theta] t1

    with t1= (v0/g)sin[theta]+ (1/g)(v0^2 sin^2 [theta]+ (gyo/2))^(1/2)
    (the positive time at which y= 0) gives

    x1= x0+ (v0^2/g)sin[theta]cos[theta]+(v0/g)sin[theta](v0^2sin^2[theta]+ (gy0/2))

    Differentiating,

    dx1/d[theta]= (v0^2/g)(cos^2[theta]- sin^2[theta])+ (v0/g)cos[theta](v0sin^2[theta]+(gy0/2))^(1/2)+ vo^3/g sin^2[theta]cos[theta](v0^2sin^2[theta] + (gy0/2))^(-1/2) = 0 at maximum x1.

    I'm with you! I don't see any "nice" way of solving that for [theta]. Do you have any reason to think that there should be?
     
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