Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

Solve the following :


A particle is projected with a speed $u$ at an angle 0 with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circle? This radius is called the radius of curvature of the curve at the point.


At highest point

$\mathrm{A}_{\text {radial }}=\frac{v 2}{R}$

$g=\frac{(u \cos \theta)^{2}}{R}$

$R=\frac{u^{2} \cos ^{2} \theta}{g}$

Leave a comment

Free Study Material