Question

A particle is projected from the ground at an angle of $\theta$ with the horizontal with an initial speed of $u$. Time after which velocity vector of the projectile is perpendicular to the initial velocity is

• A. $\frac{u}{g\sin \theta }$
• B. $\frac{u}{g\cos \theta }$
• C. $\frac{2u}{g\sin \theta }$
• D. $2u\tan \theta$

Answer 1

The correct option is A $\frac{u}{g\sin \theta }$

$\overrightarrow{u}=u_x\hat{i}+u_y\hat{j}$
$\overrightarrow{u}=u\cos \theta \hat{i}+u\sin \theta \hat{j}$
Let $v$ be the velocity which is perpendicular to $u$
At any time $t$
$v_x=u_x=u\cos \theta$ ($\because$ Horizontal component of velocity always remains constant)
$v_y=u_y-gt$
$\overrightarrow{v}=u\cos \theta \hat{i}+(u\sin \theta -gt)\hat{j}$
If $\overrightarrow{v}$ is perpendicular to $\overrightarrow{u}$ then
$\overrightarrow{v}.\overrightarrow{u}=0$
$\Rightarrow u^2{\cos }^2\theta +(u^2{\sin }^2\theta -u\sin \theta gt)=0$
$\Rightarrow u\sin \theta gt=u^2$
$t=\frac{u}{g\sin \theta }$