A body is projected at \(t = 0\) with a velocity \(10\,ms^{-1}\) at an angle of \(60^o\) with the horizontal. The radius of curvature of its trajectory at \(t=1\,s\) is \(R.\) Neglecting air resistance and taking acceleration due to gravity \(g = 10\,ms^{-2},\) the radius of \(R\) is ........ \(m\)

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(B) \(2.8\)

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\(\begin{array}{l}
at\,t = 1\\
\,{u_x} = 5,\,{u_y} = 5\sqrt 3 \\
{v_y} = 5\sqrt 3 - 10\,\,;\,\,\,{v_x} = 5\\
\tan \,\theta \, = - 2\left( {2 - \sqrt 3 } \right) \Rightarrow \theta = - {30^ \circ }\\
R = \frac{{{v^2}}}{{a \bot }} = \frac{{{{10}^2}}}{{\left( {10\,\cos \,{{30}^ \circ }} \right)}}
\end{array}\)

\(\begin{array}{l}
\,\, = \frac{{10}}{{\sqrt 3 }} \times 2\, = \frac{{20}}{{\sqrt 3 }}m\\
\frac{{{5^2} + {{\left( {10 - 5\sqrt 3 } \right)}^2}}}{{10\,\cos \theta }} = \frac{{200 - 100\sqrt 3 }}{{10 \times 0.965}} = 2.8m
\end{array}\)

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