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A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the official's boat pulls up next to the spy's boat, both boats reach the edge of a 5.0-meter waterfall. If the spy's speed is 15 m/s and the official's speed is 26 m/s how far apart will the two vessels be when they land below the water?

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Answer:

10.895 m

Explanation:

v = Velocity

g = Acceleration due to gravity = 9.81 m/s²

h = Height of waterfall = 5 m

Distance traveled by an object which has forward velocity is given by

[tex]R=\dfrac{v}{\sqrt{\dfrac{2h}{g}}}\\\Rightarrow R=\dfrac{15}{\sqrt{\dfrac{2\times 5}{9.81}}}\\\Rightarrow R=14.85681\ m[/tex]

[tex]R=\dfrac{v}{\sqrt{\dfrac{2h}{g}}}\\\Rightarrow R=\dfrac{26}{\sqrt{\dfrac{2\times 5}{9.81}}}\\\Rightarrow R=25.75181\ m[/tex]

The distance between them is 25.75181-14.85681 = 10.895 m

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