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icang [17]
3 years ago
9

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

Physics
1 answer:
leonid [27]3 years ago
3 0
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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