Question

A particle is projected from the ground at an angle of $\theta$ with the horizontal with an initial speed of $u$. Find the final velocity when it is perpendicular to the initial velocity.

• A. $u\tan \theta$
• B. $u\cot \theta$
• C. $u\sin \theta$
• D. $u\cos \theta$

Answer 1

The correct option is B $u\cot \theta$

$\overrightarrow{u}=u_x\hat{i}+u_y\hat{j}$
$\overrightarrow{u}=u\cos \theta \hat{i}+u\sin \theta \hat{j}$
Let $v$ be the velocity which is perpendicular to $u$
At any time $t$,
$v_x=u_x$ ($\therefore$ 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
$\overrightarrow{v}.\overrightarrow{u}=0$
$\Rightarrow u^2{\cos }^2\theta +(u^2{\sin }^2\theta -u\sin \theta gt)=0$
$\Rightarrow u\sin \theta gt=u^2$
$t=\frac{u}{g\sin \theta }$
$\overrightarrow{v}=u\cos \theta \hat{i}+(u\sin \theta -gt)\hat{j}......\text{(i)}$
Using $t=\frac{u}{g\sin \theta }$ in equ. $\text{(i)}$, we get
$\overrightarrow{v}=u\cos \theta \hat{i}+(u\sin \theta -g\times \frac{u}{g\sin \theta })\hat{j}$
$\overrightarrow{v}=u\cos \theta \hat{i}+\frac{u({\sin }^2\theta -1)}{\sin \theta }\hat{j}$
$\Rightarrow \overrightarrow{v}=u\cos \theta \hat{i}+(-\frac{u{\cos }^2\theta }{\sin \theta })\hat{j}$
$|\overrightarrow{v}|=\sqrt{(u\cos \theta {)}^2+\frac{u^2{\cos }^4\theta }{{\sin }^2\theta }}$
$|\overrightarrow{v}|=\sqrt{(u\cos \theta {)}^2\left(\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta }\right)}$
$|\overrightarrow{v}|=u\cot \theta$