A particle is projected with a velocity vecv=ahati+bhatj. Find the radiu of curvature of the trajectory of the particle at the (i) point of projection (ii) highest point.

A particle is projected with a velocity vecv=ahati+bhatj. Find the rad

1.	A particle is projected with a velocity vecv=ahati+bhatj. Find the radiu of curvature of the trajectory of the particle at the (i) point of projection (ii) highest point.
Answer» <P> Solution :(i) Let the angle of projection be `theta` At the point of projection `P,a_(n)=gcos theta_(0)` Hence the radius of curvature at P is `r_(p)=(v_(p)^(2))/(alpha_(n))=(v_(0)^(2))/(g cos theta_(0))` Since `TAN theta_(0)=b//a,cos theta_(0)=a/(sqrt(a^(a)+b^(2)))` Substituting `v_(0)=sqrt(a^(2)+b^(2))` and `cos theta_(0)=a/(sqrt(a^(2)+b^(2)))`, we have `R-(p)=(a^(2)+b^(2))^(3//2)g//a` (ii) At the highest position Q, the VELOCITY of the particle is `v_(Q)=v_(0)cos theta_(0)` Since it moves horizontally at highest point `Q.veca_(n)=vec(g) (_\|_ vecv)` Hence the radius of curvature at Q is `r_(Q)=(v_(Q)^(2))/(a_(n))=(v_(0)^(2)cos^(2)theta)/g` Where `v_(0)cos theta=v_(X)=a` (given) Then `r_(Q)=(a^(2))/g`