physics
posted by Anonymous .
A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the officials' boat pulls up next to the spy's boat, both reach the edge of a 4.2 m waterfall. If the spy's speed is 20 m/s and the officials' speed is 24 m/s, how far apart will the two vessels be when they land below the waterfall?

h = Vo*t + 0.5g*t^2 = 4.2 m.
20t + 4.9t^2 = 4.2
4.9t^2 + 20t  4.2 = 0
Use Quad. formula.
t1 = 0.200 s. = The spy's time.
24t + 4.9t^2 4.2 = 0
4.9t^2 + 24t  4.2 = 0
t2 = 0.169 s. = Gov. officials' time.
t1t2 = 0.200  0.169 = 0.031 s. apart. 
Correction: See recent post: Tue, 91214, 11:13 AM.
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