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?
Expert's answer

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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