# Physics

While standing on a bridge 19.1 m above the ground, you drop a stone from rest. When the stone has fallen 3.40 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.

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1. how long to fall 3.4 m ?
(1/2) (9.81) t^2 = 3.4
t = .833 seconds

how long to fall 19.1 meters ?
(1/2)(9.81) t^2 = 19.1
t = 1.97 seconds

so second rock fall for (1.97-.833) = 1.14 seconds

-19.1 = Vi t + (1/2)(-9.81) t^2
-19.1 = Vi (1.14) - (1/2)(9.81)(1.14)^2

1.14 Vi = -19.1 + 6.37

Vi = -11.2 m/s

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posted by Damon

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