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# Answer to Question #104322 in Mechanics | Relativity for james

Question #104322
An arrow is fired by an archer at a speed of 5m/s. Its path has the equation of 220st t=−meters after t seconds.

How high does the arrow go and find the speed covered by the rocket when its 3m above the ground?
1
2020-03-05T09:37:16-0500

How high the arrow goes if fired upward can be found from energy conservation principle:

"mgh=\\frac{1}{2}mv^2,\\\\\n\\space\\\\\nh=\\frac{v^2}{2g}=\\frac{5^2}{2\\cdot9.8}=1.28\\text{ m}."

Assume the archer was 5 meters above the ground. Therefore, the maximum height the arrow reaches is 6.28 m. Then the arrow falls down from rest and increases its speed. The height of 3 meters relates to 3.28 meters of free falling, therefore, the speed will be

"u=\\sqrt{2gH}=\\sqrt{2\\cdot9.8\\cdot3.28}=8.02\\text{ m\/s}."

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