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Homework Help: Free fall (check my solution im stuck) PLEASE HELP ME TODAY

  1. Sep 27, 2004 #1
    Q40. A boy stands at the edge of a bridge 20.0 m above a river and throws a stone straight down with a speed of 12.0 m/s. He throws another pebble straight upward with the same speed so that it misses the edge of the bridge on the way back down and falls into the river. For each stone find (a) the velocity as it reaches the water and (b) the average velocity while it is in flight.Note: Ignore the affects of air resistance.
    (a) X= 20 m a=9.80m/s2 V_i=12.0 m/s V_f= ?????
    V_f^2 = V_i^2 + 2a (ΔX)
    V_f^2 = (12.0 m/s)^2 + 2(9.80 m/s^2) (20 m)
    V_f^2 = (144 m^2/s^2) + (19.6 m/s^2) (20 m)
    V_f^2 = (144 m^2/s^2) + (392 m^2/s^2)
    V_f^2 = (536 m^2/s^2)
    V_f = 23.15167381 m/s
    V_f = 23 m/s
    Therefore the velocity of the stone as it reaches the water is approximately 23 m/s. Since the pebble was thrown straight upward with the same speed so that it misses the edge of the bridge on the way back down and falls into the river, I know that if the pebble is projected upwards in a vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity it has when it returns to the same height. That is, a pebble projected vertically with an upward velocity of +12 m/s will have a downward velocity of –12 m/s when it returns to that same height. So the velocity of the pebble before it reaches the water is 12 m/s. right?


    Xf = 20 m Vxf = 23 m/s Vxi=12.0 m/s ax=9.80m/s2 Xi = 0 m t=????

    V_xf = V_xi + a_x t V_avg= ΔX/Δt
    23 m/s = 12.0 m/s + (9.80 m/s^2) t V_avg= 20m/1.12244898 s
    23 m/s – 12.o m/s = (9.80 m/s^2) t V_avg= 17.81818181 m/s
    11 m/s = (9.80 m/s^2) t V_avg= 17.81 m/s
    t = 11 m/s / 9.80 m/s^2

    t = 1.12244898 s
    t = 1.12 s


    X_f = 20 m V_xf = 12.0 m/s V_xi=12.0 m/s a_x=9.80m/s2 X_i = 0 m t=????

    V_xf = V_xi + a_x t V_avg= ΔX/Δt
    12.0 m/s = 12.0 m/s + (9.80 m/s^2) t V_avg= 20 m/ o s
    12.0 m/s – 12.0 m/s= (9.80 m/s^2) t V_avg=
    0 m/s = (9.80 m/s^2) t
    0 m/s / 9.80 m/s2 = t
    t = 0 s
    how can t be zero and how am i suppose tocalculate velocity now PLEASE HELP ME
  2. jcsd
  3. Sep 27, 2004 #2
    Can Someone Pleasse Help Me Im Gonna Be Staying Up All Nihgt Trying Tofigure This Out
  4. Sep 27, 2004 #3


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    Homework Helper

    No. Its downward velocity AT THE HEIGHT OF THE BRIDGE is 12 m/s. You took the downward direction positive. The situation with the pebble is the same as it was for the stone. Both have the same velocity at the same height, so the final velocity must be the same, too.

    Yes. You know ΔX and you know some formulas for constant acceleration:
    vf=vi+a*t , xf=xi+vi*t+0.5*a*t^2 or xf-xi=0.5 (vf+vi)*t.

    vf=23 m/s, vi=-12 m/s (it was thrown upward, and you took downward as the positive direction).

    Use xf-xi=0.5 (vf+vi)*t to find t. (20=0.5*(-12+23)t ---->t=3.636 s, V_avg= ΔX/Δt=20/3.636=5.5 m/s)

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