Question

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

• A. $\frac{u}{2}\sqrt{1+4{\cos }^2\theta }$
• B. $\frac{u}{2}\sqrt{1+2{\cos }^2\theta }$
• C. $\frac{u}{2}\sqrt{1+3{\cos }^2\theta }$
• D. $\frac{u}{2}\sqrt{1+{\cos }^2\theta }$

Answer 1

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

We know that,
$$\text{Average velocity (v)}=\frac{\text{Total displacement (S)}}{\text{Total time (t)}}.....\left(1\right)$$
Diagram for projectile motion:



From point $\text{O}$ to $\text{B}$, the displacement $S$ is,
$S=\sqrt{H^2+{\left(\frac{R}{2}\right)}^2}$

Substituting the height $\left(H\right)$ of projectile at its highest point and horizontal distance $\left(\frac{R}{2}\right)$ covered till this point. we get,

$S=\sqrt{{\left(\frac{u^2{\sin }^2\theta }{2g}\right)}^2+{\left(\frac{u^2\sin 2\theta }{2g}\right)}^2}$

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

Time taken to reach highest point is $t$, which is equal to the half of time of flight $T$,

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

Substituting the values in equation $\left(1\right)$, we get

$<v>=\frac{\frac{u^2}{2g}\sqrt{({\sin }^2\theta {)}^2+4{\sin }^2\theta {\cos }^2\theta }}{\frac{u\sin \theta }{g}}$

$\Rightarrow <v>=\frac{u}{2}\sqrt{\frac{({\sin }^2\theta \times {\sin }^2\theta )+4{\sin }^2\theta {\cos }^2\theta }{{\sin }^2\theta }}$

$\Rightarrow <v>=\frac{u}{2}\sqrt{{\sin }^2\theta +4{\cos }^2\theta }$

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

$\therefore <v>=\frac{u}{2}\sqrt{1+3{\cos }^2\theta }$

Hence, option (c) is correct answer.