Question

lf a particle is projected with speed $u$ from ground at an angle $\theta$ with horizontal, then radius of curvature of a point where velocity vector is perpendicular to initial velocity vector is given by :

• A. $\frac{u^2{\cos }^2\theta }{g}$
• B. $\frac{u^2{\cot }^2\theta }{gsin\theta }$
• C. $\frac{u^2}{g}$
• D. $\frac{u^2{\tan }^2\theta }{gcos\theta }$

Answer 1

The correct option is B $\frac{u^2{\cot }^2\theta }{gsin\theta }$

Required, $\overrightarrow{v}.\overrightarrow{u}=0$

or, $(u_x\hat{i}+(u_y-gt\left)\hat{j}\right).(u_x\hat{i}+u_y\hat{j})=0$

which gives, $u_x^2+u_y^2-u_{y}gt=0$

or, $u^2{\cos }^2\theta +u^2{\sin }^2\theta -ugt.\sin \theta =0$

Thus, $t=\frac{u}{g\sin \theta }$

Now, trajectory of projectile is given as:

$x=u\cos \theta .t$

$y=u\sin \theta .t-\frac{1}{2}g{t}^2=x\tan \theta -\frac{1}{2}g.\frac{x^2}{u^2{\cos }^2\theta }$
Thus we have,

$R=\frac{[1+(\frac{dy}{dx}){}^2{]}^{3/2}}{\left|\frac{d^{2}y}{d{x}^2}\right|}$
$=\frac{[1+(\tan \theta -\frac{gx}{u^2{\cos }^2\theta }){}^2{]}^{3/2}}{\left|\frac{-g}{u^2{\cos }^2\theta }\right|}$
Now, at $t=\frac{u}{g\sin \theta }$, $x=u\cos \theta .\frac{u}{g\sin \theta }=\frac{u^2}{g}cot\theta$

Put this in equation for R, we get:
$R=\frac{[1+(\tan \theta -\frac{cot\theta }{{\cos }^2\theta }){}^2{]}^{3/2}}{\left|\frac{-g}{u^2{\cos }^2\theta }\right|}$
$=\frac{[1+(\frac{{\sin }^2\theta -1}{\sin \theta \cos \theta }){}^2{]}^{3/2}}{\left|\frac{-g}{u^2{\cos }^2\theta }\right|}$
$=\frac{[1+\cot {}^2\theta {]}^{3/2}}{\left|\frac{-g}{u^2{\cos }^2\theta }\right|}=\frac{u^2}{g}\frac{{\cot }^2\theta }{\sin \theta }$