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Homework Help: Projectile Motion and linear motion problem

  1. Jan 6, 2008 #1
    A projectile is launched at an angle α from the edge of a cliff of height H above sea level. If it falls
    into the sea a distance D from the base of the cliff, prove that its maximum height above sea level is

    H + [(Dtan α)^2]/4(H + Dtan α)

    Newtons equations of linear motion in the i/j plane.

    Well, I understand that the max height is at a height H + (something) with that something being Sjmax, when Vj=0.

    Ive got my Sjmax = (usin α)^2/ 2g

    This is where i get a little stuck... Ive tried a couple of things but the fact that its a cliff is making my reasoning wrong, if it wasnt a cliff, you could also say that the max vertical height is obtained when the horizontal displacement is D/2, the christmas break has made the old mind a little blunt... Any hints? Thanks so much in advance.
  2. jcsd
  3. Jan 6, 2008 #2
    You have to use the distance D, to find u.

    In what axis is D, and when the projectile reaches the sea?
  4. Jan 6, 2008 #3
    Ya, I can see you need to find an equation for u, because thats the unknown that isnt in the answer.

    D is in the i plane, so do I find equations for when Sj= -H, and Si=D?
  5. Jan 6, 2008 #4
    Yes! That's the way! :smile:
  6. Jun 28, 2010 #5
    Not the most elegant solution but here is a way:

    lets split the trajectory into two: one part where it reaches the same height as the starting point, and second part where it goes into the sea. Lets use H1, D1 and H2, D2 to denote the heights and distances of each segment:

    H2 = H
    D1 + D2 = D

    The solution we are looking for is the height of the first parabola, H1. Let's use u to denote the initial velocity of projection. By solving the equations of motion in the i and j planes, we can get the following:

    D1 = 2u2 sin[tex]\alpha[/tex] cos[tex]\alpha[/tex] / g

    H1 = u2sin2 [tex]\alpha[/tex]/2g

    for the second projectile, lets calculate the time of travel using the i direction:

    D2 = D-D1 = ucos[tex]\alpha[/tex] * t

    t = (D-D1)/ucos [tex]\alpha[/tex] =D/ucos[tex]\alpha[/tex] - 2usin[tex]\alpha[/tex]/g = Dsec[tex]\alpha[/tex]/u - 2usin[tex]\alpha[/tex]/g

    use this in the equation for j direction; note here the initial velocity will be usin[tex]\alpha[/tex] pointing downwards:

    H = usin[tex]\alpha[/tex] *t + 1/2*g*t2 = Dtan[tex]\alpha[/tex] - 2u2sin2[tex]\alpha[/tex]/g + g/2 * [ D2sec2[tex]\alpha[/tex]/u2 +4u2sin2[tex]\alpha[/tex]/g2 - 4Dtan[tex]\alpha[/tex]/g]
    = Dtan[tex]\alpha[/tex] - 2u2sin2[tex]\alpha[/tex]/g + gD2sec2[tex]\alpha[/tex]/2u2 +2u2sin2[tex]\alpha[/tex]/g - 2Dtan[tex]\alpha[/tex]
    = gD2sec2[tex]\alpha[/tex]/2u2 - Dtan[tex]\alpha[/tex]
    Rearranging terms:
    so, u2 = gD2sec2[tex]\alpha[/tex]/[2(H+Dtan[tex]\alpha[/tex])]


    H1 = u2sin2[tex]\alpha[/tex]/2g = gD2sec2[tex]\alpha[/tex] * sin2[tex]\alpha[/tex] / [2g *2(H + Dtan[tex]\alpha[/tex])]
    = D2tan2[tex]\alpha[/tex]/[4(H + Dtan[tex]\alpha[/tex])] // sin [tex]\alpha[/tex] * sec [tex]\alpha[/tex] = sin[tex]\alpha[/tex] / cos[tex]\alpha[/tex] = tan[tex]\alpha[/tex]

    total height = H1 + H2 = H + (Dtan[tex]\alpha[/tex]) 2 / 4 (H + Dtan[tex]\alpha[/tex]) <------ solved!

    I tried my best to maintain clarity but let me know if something is not clear!
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