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A particleis projected witha velocity bar(v) = ahat(i) + bhat(j) . Findthe radiusof curvature of thetrajectory of the particle at (i) pointof projection (ii) highestpoint . |
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Answer» <P> Solution :(i) Letthe angleof projection be `theta` .At thepoint of projection `P, a_(n) = g cos theta_(0)` . Hencethe radiusof curvatureatP 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 thehighestposition Q, thevelocityof the PARTICLE is `v_(Q) = v_(0) cos theta_(0)` Sinceit moves HORIZONTALLY athighestpoint`Q.bar(a_(n)) = bar(g) (BOT bar(v))` Hencethe 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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