Question

solve this asap......​

Answer 1

➳GIVEN:-

Particle is projected from the ground with intial speed of V.

At an angle $\theta$ with horizontal.

➳FIND:-

Average velocity of particle between between it's points of projection and heighest point of trajectory = ?

➳SOLUTION:-

we know,

$\sf⎆Average \: Velocity = \frac{Displacement}{Time}$

$\sf⎆V_{av} = \frac{ \sqrt{ {H}^{2} + \frac{ {R}^{2} }{4} } }{ \frac{T}{2} }$

where, H = $\sf max^m\:height$

$\sf ➲ H = \frac{V^2sin^2\theta}{2g}$

$\sf ➲ Time \: t = time \: of \: flight = \frac{2Vsin\theta}{g}$

$\sf☄Hence, \boxed{ \sf v_{av} = \frac{v}{2} \sqrt{1 + 3 { \cos }^{2} \theta } }$

Answer 2

Answer:

B is ur answer.

Pls make my answer brainliest and thank my answers.