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Q.

A particle is projected from the ground with an initial speed of v at an angle of projection $θ$ . The average velocity of the particle between its time of projection and time it reaches highest point of trajectory is

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a

$\frac{v}{2} \sqrt{1 + 2 \sin^{2} θ}$

b

v cos $θ$

c

$\frac{v}{2} \sqrt{1 + 3 \cos^{2} θ}$

d

$\frac{v}{2} \sqrt{1 + 2 \cos^{2} θ}$

answer is C.

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Detailed Solution

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$\begin{array}{l} time = t_{a} = \frac{Vsin θ}{g} \\ s = \sqrt{h^{2} + \frac{R^{2}}{4}} \end{array} \begin{array}{l} = \sqrt{{(\frac{V^{2} \sin^{2} θ}{2 g})}^{2} + \frac{1}{4} {(\frac{V^{2} \sin 2 θ}{g})}^{2}} \\ = \frac{V^{2} \sin θ}{2 g} \sqrt{1 + 3 \cos^{2} θ} \\ \begin{array}{ll} V_{av} = \frac{s}{t} = & = \frac{(\frac{V^{2} \sin θ}{2 g})}{\frac{Vsin θ}{g}} \sqrt{1 + 3 \cos^{2} θ} \end{array} = \frac{V}{2} \sqrt{1 + 3 \cos^{2} θ} \end{array}$

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