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

Explanation

(i) Let the angle of projection be `theta`, At the point of projection, `p,a_(n)=g cos theta_(0)`. Hence, the radius of curvature at P is `r_(P)=(v_(P)^(2))/(a_(n))=(v_(0)^(2))/(g cos theta_(0))`

Since, `tan theta_(0)=(b)/(a),cos theta_(0)=(a)/(sqrt(a^(2)+b^(2))`' substituting `v_(0)=sqrt(a^(2)+b^(2))` and `cos theta=(a)/(sqrt(a^(2)+b^(2))`
We have, `r_(P)=((a^(2)+b^(2)^(3//2))/(ga))`
At the highest position Q, the velocity of the particle is `v_(Q)=v_(0)cos theta_(0)`. since, it moves horizantlly at the highest point `Q,veca_(n)=vecg(botvecv)`. 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)`