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, $y=\frac{u^2{\sin }^2\theta }{2g}$
$x=\frac{R}{2}=\frac{u^2\sin 2\theta }{2g}$
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 $=\frac{\bar{r}}{t}=u\cos \theta \hat{i}+\left(u\sin \theta -\frac{gt}{2}\right)\hat{j}$
$\left(V_{AV}\right)=\sqrt{u^2{\cos }^2\theta +u^2{\sin }^2\theta -ugt\sin \theta +\frac{g^2{t}^2}{4}}$
$=\sqrt{u^2+\frac{g^2{t}^2}{4}-ugt\sin \theta }$
For $\left(V_{AV}\right)>u$
$\Rightarrow \frac{g^2{t}^2}{4}>ugt\sin \theta$ or $t<\left(\frac{4u\sin \theta }{g}\right)$