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A particle is projected with a velocit...

A particle is projected with a velocity `bar(v) = ahat(i) + bhat(j)` . Find the radius of curvature of the trajectory of the particle at (i) point of projection (ii) highest point .

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(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))/(alpha_(n)) = (v_(0)^(2))/(g cos theta_(0))`
Since `tan theta _(b) = b//a , 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.bar(a_(n)) = bar(g) (bot bar(v))`
Hence the radius of curvature at Q is .
`t_(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)`
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