leyla7
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An object is 45 m above the ground when it is dropped. How fast is the object going just before it hits the ground?

Verified Answer :

SapphireMoon

This is a kinematics question.
\(v^{2} = v_{0} ^{2} + 2g(y - y_{0}) \\ v^{2} = 0 + 2(-9.8)(0 - 45) \\ v^{2} = 882 \\ v = \sqrt{882} = 29.7 m/s\)

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