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The distance (not horizontal nor vertical) of a rock traveled in a projectile

  1. Dec 9, 2007 #1
    1. The problem statement, all variables and given/known data
    A small rock of mass m is kicked from the edge of a perfectly vertical cliff of height h, giving
    it an initial velocity ~vo that’s purely horizontal. The rock lands at a distance d away from
    the bottom of the cliff. The ground there is assumed horizontal and we neglect air resistance.
    Derive the distance traveled by the rock in terms of h and d only.


    2. Relevant equations
    (h) y=-gt^2/2
    (d) x=v0t
    Using the two parametric equations above and the arc length formula for parametric equations.
    Integrate[Sqrt[(y')^2+(x')^2],t,0,x/v0]

    3. The attempt at a solution
    We could not solve the integral
    and the professor did not allow us to have vo and t in the final answer!!!

    Help...it's due tmr....
     
  2. jcsd
  3. Dec 9, 2007 #2

    rl.bhat

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    Distance traveled = Integration(v*dt) = Int[sqrt{(gt)^2 + vo^2}]*dt
    s = Int[vo*sqrt{(gt/vo)^2 + 1}]*dt = vo*Int.[sqrt{(gt/vo)^2 + 1}]*dt
    Let gt/vo = 2*1/2*gt^2/vo*t = 2h/d = tan(theta)
    (g/vo)*dt = sec^2(theta)*d(theta) and dt = [sec^2(theta)*d(theta)]vo/g. Substitute in s, we get
    s = (vo^2/g)*Int[sec^3(theta)*d(theta). If you solve this by integration by parts, we get
    s = (d^2/4h)*[sqrt{1 + 4h^2/d^2}*2h/d + log(sqrt{1 + 4h^2/d^2}+2h/d )]
     
    Last edited: Dec 10, 2007
  4. Dec 10, 2007 #3

    robphy

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    So the distance traveled has units of length^2 ?

    It seems that the OP is looking for the arc-length along a segment of a parabola.
    Can you write the integrand in simplest form?
     
    Last edited: Dec 10, 2007
  5. Dec 12, 2007 #4

    rl.bhat

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    Distance traveled = Integration(v*dt) = Int[sqrt{(gt)^2 + vo^2}]*dt
    s = Int[vo*sqrt{(gt/vo)^2 + 1}]*dt = vo*Int.[sqrt{(gt/vo)^2 + 1}]*dt
    Let gt/vo = 2*1/2*gt^2/vo*t = 2h/d = tan(theta)
    (g/vo)*dt = sec^2(theta)*d(theta) and dt = [sec^2(theta)*d(theta)]vo/g. Substitute in s, we get
    s = (vo^2/g)*Int[sec^3(theta)*d(theta). If you solve this by integration by parts, we get
    s = (d^2/4h)*[sqrt{1 + 4h^2/d^2}*2h/d + log(sqrt{1 + 4h^2/d^2}+2h/d )]
     
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