Question

A body is projected from a point on ground making an angle of 60^o with horizontal. If radius of curvature of its trajectory at the top most point is 10 m, then radius of curvature at the point of projection is

Difficult

Solution

If u be the velocity of projection, $α$ be the angle of projection and $ρ$ be the radius of curvature at the highest point, then
$\frac{{(u \cos α)}^{2}}{ρ} = g \Rightarrow ρ = \frac{u^{2} \cos^{2} α}{g}$
At point A, $\frac{u^{2}}{ρ^{1}} = g \cos α \Rightarrow g \cos α \Rightarrow ρ^{1} = (\frac{u^{2}}{g}) \frac{1}{\cos α} = \frac{ρ}{\cos^{2} α} . \frac{1}{\cos α} = \frac{ρ}{\cos^{3} α}$
$∴ ρ^{1} = \frac{10}{{(\cos 60^{o})}^{3}} m = 80 m$

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