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Stone thrown question

  1. Feb 19, 2007 #1
    A stone is thrown vertically downward from a 2.00x10^2m high cliff at an initial velocity of 5.00m/s. (a) How long does it take for the stone to reach the base of the cliff? (b) What is the stones final velocity?

    a) d = vi(t)+1/2a(t)^2
    0= -200 + (5)t + (4.910^2
    4.91(t)^2 + (5)t -200
    t= -5+- sqaureroot sign (5)^2-4(4.91)(-200)
    -5+- squareroot sign 25+3953 / 9.81

    so i used the quadratic fromula

    b) vf^2=vi^2+2ad
    vf^2 = 3949
    vf= 63m/s
  2. jcsd
  3. Feb 19, 2007 #2
    If you're just looking for a homework check station, you at least need to explain some uncertainty in reasoning. I can't speak for everyone here, esp the PF mentors.

    For this problem, 63m/s as a dbl check should equate to the velocity to that gained by freefall in 200m less the distance from the initial velocity, then added to the initial velocity. If we take 6*5 and subtract from 200= 170 for the distance of the grav assist. We know that v=sqrt(2*a*y)
    =sqrt (2*9.81*170), I get 57.75. Add to that 5, its close.
  4. Feb 19, 2007 #3
    Yeah, its correct.
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