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Home Q & A Board Science-Related Homework Help Physics
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Tonietn8 Tonietn8
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10 years ago
While standing on a bridge 15.4 m above the ground, you drop a stone from rest. When the stone has fallen 3.10 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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irinairina
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10 years ago
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10 years ago
The two stones m and M travel the same distance H = 15.4 m.  

When m has dropped S = 1/2 gt^2 = 3.1 m, the time of drop is t = sqrt(2S/g) = ? seconds and the remaining distance to fall is D = H - S = 12.3 meters = uT + 1/2 gT^2, which will take T seconds to go that D remaining fall.  u = gt the velocity of m after falling that t seconds to reach S = 3.1 m.  Solve for u.

Solve for T seconds using the quadratic 0 = 1/2 gT^2 + uT - 12.3.

The mass M must fall H = UT + 1/2 gT^2 in that remaining T seconds of fall by m.  U = ? the initial velocity M must have to catch up with m upon impact.  Solve for U.  (H - 1/2 gT^2)/T = U = ? mps the answer you are looking for.  You can do the math.
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