Question

A particle is projected from ground at an angle θ with horizontal with speed u. The ratio of radius of curvature of its trajectory at point of projection to radius of curvature at maximum height is?
Options are
A)1/sin^2θ cosθ B)cos^2θ
C)1/sin^3θ D)1/cos^3θ

Answer 1

$x(t)=u\ Cos\theta\ t,\ \ y(t)=u\ Sin\theta\ t - \frac{1}{2} g t^2 \\\\ Radius\ of\ curvature\ R=| \frac{[(x'(t))^2+(y'(t))^2]^{3/2}}{x'(t)*y"(t)-x"(t)*y'(t)} |\\\\ x'(t)=u\ Cos\theta,\ \ x''(t)= 0\\\\ y'(t)=u\ Sin\theta-gt\\ y''(t)= -g\\\\ at t=0, x'(0)=u\ Cos\theta,\ y'(0)=u Sin\theta,\ x''(0)= 0,\ y''(0)=- g\\\\ at\ t=\frac{uSin\theta}{g},\ x'=u\ cos\theta,\ y'=uSin\theta,\ y''=-g$


$Substituting\ the\ values\ we\ get\\\\ R(t=0) = \frac{u^2}{gCos\theta}\\\\ R(t=uSin\theta/g)=\frac{u^2Cos^2\theta}{g}\\\\Ratio=1/Cos^3\theta$

Answer 2

Answer:

radius of curvature = [u²cos²(Theeta)]/g