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A 2.6 kg rock is dropped from a height of 10 m. With what speed will it strike the ground. Ignore air resistance. Solve using conservation of energy (start with Ebefore = Eafter).​

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Answer:

[tex]mgh = \frac{1}{2} m {v}^{2} \\ v = \sqrt{2gh} \\ = \sqrt{2 \times 9.8 \times 10} \: m. {s}^{ - 1} [/tex]

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