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A particle is...

Question

A particle is projected with a speed $v$ at an angle $θ$ with the horizontal. Find the mean power delivered by gravity during the ascent of the particle

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Solution

The magnitude of mean power for $= ¯ ¯¯ ¯ P = \frac{1}{t_{0}} \int_{0}^{t_{0}} m \to g . {\to v}_{y} d t = \frac{1}{t_{0}} \int_{0}^{t_{0}} m g . v_{y}^{'} d t [v_{y}^{'} = (v_{y} - g t)]$
$\Rightarrow ¯ ¯¯ ¯ P = \frac{1}{t_{0}} \int_{0}^{t_{0}} m g (v_{y} - g t) d t; t_{0} = \frac{v sin θ}{g} \Rightarrow ¯ ¯¯ ¯ P = \frac{m g v_{y}}{2} = \frac{m g v sin θ}{2}$

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