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The Empire State Building is about 381m high. If you threw a baseball downward at 25 m/s from the top of the building how fast would it be traveling just before it landed?

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Answer:

Explanation:

You can use that [tex]v = v_{o}+gt[/tex] or [tex]v^{2}=v_{o}^{2}+2gh=(25)^{2}+(2)(10)(381)=8245 \rightarrow v \approx 90.8[/tex] m/s.

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