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2d motion question

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data

    wiley e coyote is out once more to try to capture the elusive roadrunner. the coyote wears a pair of acme jet powered roller skates, which provide a constant horizontal acceleration ac. the coyote starts off at rest D meters from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff. the cliff is H meters above the canyon floor. given [ac, D, H], determine:

    if the roadrunner moves at constant speed, what should the roadrunners min speed be in order to reach the cliff before the coyote?

    as usual the roadrunner is saved by making a sudden turn at the cliff. determine where the coyote lands in the canyon. (assume skates still on)

    the velocity of the coyote just before hitting the canyon floor

    2. Relevant equations

    vx=v0x+at
    x=x0+v0x*t+1/2*a*t^2
    vx^2=v0x^2 + 2a(x - x0)

    and y equivilents



    3. The attempt at a solution


    for a, i solved for time(t1) as t1= sqrt(2D/ac), which i then use for velocity of roadrunner(Vr) as Vr= D/sqrt(2D/ac)

    for b & c, i used info from solving things in a as info to parameters:

    from D to bottom of cliff

    x0= D
    x= D2
    Vx0= acsqrt(2D/ac)
    Vx= ?
    a = ac

    y0=0
    y= H
    Vy0 = 0
    Vy = ?
    a = g

    t=t2

    trying to use my y-based equations to solve for t2 results in them collapsing

    Vy= 0 + gt2
    Vy^2 = 0^2 + 2gH

    g^2*(t2)^2 = 2gH

    H = (g*t2^2)/2

    (g*t2^2)/2 = 0 + 0 + 1/2*g*(t2)^2

    everything cancels :<
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 18, 2012 #2
    The trajectory will have two distinct parts. Find the velocity at the end of the first part. The second part, due to the jets still on, will be accelerated both horizontally and vertically. Use the final velocity of part one to find what is required.
     
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