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

\frac{v}{2} \sqrt{1 + 2 {cos}^{2} θ}

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\frac{v}{2} \sqrt{1 + {cos}^{2} θ}

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\frac{v}{2} \sqrt{1 + 3 {cos}^{2} θ}

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v cos θ

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Solution

The correct option is C $\frac{v}{2} \sqrt{1 + 3 {cos}^{2} θ}$
For highest point-

$T = (t o t a l - t i m e) / 2 = \frac{v sin θ}{g}$

$x = v cos θ \times T = \frac{v^{2} sin θ cos θ}{g}$

$y = \frac{v^{2} {sin}^{2} θ}{2 g}$

displacement $s = \sqrt{x^{2} + y^{2}} = \frac{v^{2}}{g} \sqrt{{sin}^{2} θ {cos}^{2} θ + \frac{{sin}^{4} θ}{4}}$

$v_{a v} = \frac{s}{T} = v \sqrt{{cos}^{2} θ + \frac{{sin}^{2} θ}{4}}$

$⟹ v_{a v} = \frac{v}{2} \sqrt{3 {cos}^{2} θ + 1}$

Answer-(C)

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