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Physics-

General

Easy

Question

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

$\frac{v}{2} \sqrt{1 + 2 \cos^{2} θ}$
$\frac{v}{2} \sqrt{1 + \cos^{2} θ}$
$\frac{v}{2} \sqrt{1 + 3 \cos^{2} θ}$
v cos q

The correct answer is: $fraction numerator v over denominator 2 end fraction square root of 1 plus 3 cos to the power of 2 end exponent invisible function application blank theta end root$

$fraction numerator R over denominator 2 end fraction equals fraction numerator v to the power of 2 end exponent sin invisible function application 2 theta over denominator 2 g end fraction$
displacement = $square root of H to the power of 2 end exponent plus fraction numerator R to the power of 2 end exponent over denominator 4 end fraction end root$
v_avg = $fraction numerator D i s p l a c m e n t over denominator t end fraction$
$equals fraction numerator v over denominator 2 end fraction square root of 1 plus 3 cos to the power of 2 end exponent invisible function application theta end root$

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