physics

While standing on a bridge 18.5 m above the ground, you drop a stone from rest. When the stone has fallen 2.80 m, you throw a second stone straight down. What minimal initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction.

asked by shaknocka
  1. d = 0 + 0.5*9.8t^2 = 2.8m.

    4.9t^2 = 2.8,
    t^2 = 2.8 / 4.9 = 0.57,
    t = 0.76s to drop 2.8m.

    d = 0.5 * 9.8t^2 = 18.5m,
    4.9t^2 = 18.5,
    t^2 = 18.5 / 4.9 = 3.78,
    t1 = sqrt3.78 = 1.94s. = Time in flight
    for 1st stone.

    t2 = 1.94 - 0.76 = 1.18s. = Time in flight for 2nd stone.

    d=Vi*(1.18) + 0.5*9.8*(1.18)^2=18.5,
    1.18Vi + 6.82 = 18,5,
    1.18Vi = 18.5 - 6.82 = 11.68,
    Vi = 11.68 / 1.18 = 9.9m/s.

    posted by Henry

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