Question

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

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

Answer 1

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

$\text{Step 1: Height, Range and Time of Projectile}$ $\text{[Ref. Fig. 1]}$

Height of the Projectile: $H=\frac{u^2{\sin }^2\theta }{2g}$

Range ofprojectile: $R=\frac{u^2\sin 2\theta }{g}$

Time of flight : $T=\frac{2u\sin \theta }{g}$

$\text{Step 2: Average velocity calculation}$

The magnitude of average velocity of the projectile from start to its peak is

$|\overrightarrow{V_{AV}}|=\frac{\text{|Displacement (OP)|}}{\text{time to reach OP}}$ $....\left(1\right)$

Time to reach $OP{t}_1=\frac{T}{2}=\frac{u\sin \theta }{g}$ $....\left(2\right)$

$\text{Step 3: Displacement OP}$ $\text{[Ref. Fig. 2]}$

At Point $P$, horizontal displacement $=\frac{R}{2}$

$OP=\sqrt{H^2+{\left(\frac{R}{2}\right)}^2}$

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

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

$=\frac{u^2\sin \theta }{2g}\sqrt{1+3{\cos }^2\theta }$ $....\left(3\right)$

$\text{Step 4: Solving above equations}$

Using equation (1), (2) and (3)

$|\overrightarrow{V_{AV}}|=\frac{u^2\sin \theta }{2g}\frac{\sqrt{1+3{\cos }^2\theta }}{\frac{u\sin \theta }{g}}$

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

Hence, the correct option is C.