Question

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

Answer 1

application of oblique projectile

Answer 2

Answer:

Option (C) is correct.

Explanation:

The initial velocity of the particle projected is $v_{i}$

The angle of the projectile is $\theta$

The component of the initial velocity is

$v_{i}=v \cos \theta \hat i+v \sin \theta \hat j$

The velocity at the peak of the projectile

The vertical velocity will be zero at the peak only the horizontal velocity will be there

$v_{f}=v\cos \theta\hat i$

The average velocity is,

$v_{avg}=\dfrac{v_{f}+v_{i}}{2}\\v_{avg}=\dfrac{v\cos \theta\hat i+v \cos \theta\hat i+v\sin \theta\hat j}{2}\\v_{avg}=\dfrac{2v \cos \theta \hat i+v\sin \theta \hat j}{2}$

The magnitude of the average velocity is,

$v_{avg}=\sqrt{\dfrac{(2v\cos \theta)^2}{2^2}+\dfrac{(2v\sin \theta)^2}{2^2}}\\\\v_{avg}=\dfrac{v}{2}\sqrt{4cos^2 \theta+\sin^2\theta}\\\\v_{avg}=\dfrac{v}{2}\sqrt{1+3\cos^2 \theta}$

So option (C) is correct.