Question

Statement 1: A particle is projected from origin under gravity with the angle of projection $\theta=\tan ^{-1}(2)$. The radius vector locating highest point is at an angle of elevation $\frac{\pi}{4}$
Statement-2: The ratio of average velocity to the velocity of projection is greater than unity for the particle projected under gravity at some angle from horizontal

• A. Statement-1 is true, Statement-2 is false
• B. Statement-1 is false, Statement-2 is true
• C. Statement-1 is true, Statement-2 is true and Statement-2 is a correct explanation for Statement-1
• D. Statement-1 is true, Statement-2 is true but Statement-2 is not a correct explanation for Statement-1

Answer 1

Answer: (D)

Statement-1 is true, Statement-2 is true but Statement- 2 is not a correct explanation for Statement-1.
At the highest point, $y=\frac{u^{2} \sin ^{2} \theta}{2g}$
$x=\frac{R}{2}=\frac{u^{2} \sin 2 \theta}{2 g}$
The radius vector locating highest point, $\tan\, \alpha=\frac{y}{x}$
$=\frac{\sin ^{2} \theta}{\sin 2 \theta}=\frac{\sin ^{2} \theta}{2 \sin \theta \cos \theta}=\frac{\tan \theta}{2}$
$=\frac{\tan \left(\tan ^{-1} 2\right)}{2}=1$ or $\alpha=\frac{\pi}{4}$
If velocity of projection be $u$ , then position vector at any time $t$
$\vec{r}=(u \cos \theta . t) \hat{i}+\left(u \sin \theta \cdot t-\frac{1}{2} g t^{2}\right) \hat{j}$
Average velocity $=\frac{\bar{r}}{t}=u \cos \theta \hat{i}+\left(u \sin \theta-\frac{g t}{2}\right) \hat{j}$
$\left(V_{A V}\right)= \sqrt{u^{2} \cos ^{2} \theta+u^{2} \sin ^{2} \theta-u g t \sin \theta+\frac{g^{2} t^{2}}{4}} $
$=\sqrt{u^{2}+\frac{g^{2} t^{2}}{4}-u g t \,\sin\, \theta} $
For $\left(V_{A V}\right)>\,u$
$\Rightarrow \frac{g^{2} t^{2}}{4}>\,u g t \sin \theta$ or $t<\left(\frac{4 u \sin \theta}{g}\right)$