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

Correct answer: C. (v)/(2) 1+3 ^2 θ From figure,average velocity, v_ av = H^2+R^2 / 4 T / 2 (i) Here, H=(u^2 ^2 θ)/(2 g) R=(u^2 2 θ)/(g) and T=(2 u θ)/(g) Putting these value in (i), we get v_ av =(v)/(2) 1+3 ^2 θ

AIIMS 2013

Exam: AIIMS
Year: 2013
Subject: Physics
Chapter: Motion in Two Dimensions

AIIMS 2013 · Physics · Motion in Two Dimensions

A particle is projected from the ground with an initial speed of '...

A particle is projected from the ground with an initial speed of '

v

' at angle

\theta

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

A particle is projected from the ground with an initial speed of ' v ' at angle θ with horizontal. The average velocity

Options

A $\frac{v}{2} \sqrt{1+2 \cos ^2 \theta}$
B $\frac{v}{2} \sqrt{1+\cos ^2 \theta}$
C $\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}$
D $v \cos \theta$

Correct answer

C. $\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}$

Step-by-step solution

From figure,average velocity, A particle is projected from the ground with an initial speed of ' v ' at angle θ with horizontal. The average velocity

v_{\mathrm{av}}=\frac{\sqrt{H^2+R^2 / 4}}{T / 2}

\ldots(\mathrm{i})

Here,

H=\frac{u^2 \sin ^2 \theta}{2 g}

R=\frac{u^2 \sin 2 \theta}{g}

and

T=\frac{2 u \sin \theta}{g}

Putting these value in (i), we get

v_{\mathrm{av}}=\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}

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