Question

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

• A. u cos $\theta$
• B. $\frac{u}{2}\sqrt{1+co{s}^2\theta }$
• C. $\frac{u}{2}\sqrt{1+2co{s}^2\theta }$
• D. $\frac{u}{2}\sqrt{1+3co{s}^2\theta }$

Answer 1

The correct option is D $\frac{u}{2}\sqrt{1+3co{s}^2\theta }$

The angle of the projectile is $\theta$

The initial velocity of the particle protected is $V_i$

$\overrightarrow{u_i}=\overrightarrow{v_i}=v\cos \theta \hat{i}+v\sin \theta \hat{j}$

The final velocity of the particle is

$\overrightarrow{u_f}=\overrightarrow{v_f}=v\cos \theta \hat{i}$ ($\because$ The vertical velocity will be zero at the peak only horizontal velocity will be there)

Average velocity is ,

${\overrightarrow{u}}_{avg}=\frac{\overrightarrow{v_i}+\overrightarrow{v_f}}{2}$

$=\frac{v\cos \theta \hat{i}+v\sin \theta \hat{j}+v\cos \theta \hat{i}}{2}$

$=\frac{2v\cos \theta \hat{i}+v\sin \theta \hat{j}}{2}$

The magnitude of average velocity is,

$a_{avg}=v_{avg}=\sqrt{{\left(\frac{2v\cos \theta \hat{i}+v\sin \theta \hat{j}}{2}\right)}^2}$

$=\frac{u}{2}\sqrt{4{\cos }^2\theta +{\sin }^2\theta }$

$=\frac{u}{2}\sqrt{3{\cos }^2\theta +({\cos }^2\theta +{\sin }^2\theta )}$

$=\frac{u}{2}\sqrt{3{\cos }^2\theta +1}$ $\because ({\sin }^2\theta +{\cos }^2\theta =1)$

$=\frac{u}{2}\sqrt{3{\cos }^2\theta +1}$

$\therefore$ correct option is (D)