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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 ........ $m$
$10.3$
$2.8$
$2.5$
$5.1$
Solution
$\begin{array}{l}
at\,t = 1\\
\,{u_x} = 5,\,{u_y} = 5\sqrt 3 \\
{v_y} = 5\sqrt 3 – 10\,\,;\,\,\,{v_x} = 5\\
\tan \,\theta \, = – 2\left( {2 – \sqrt 3 } \right) \Rightarrow \theta = – {30^ \circ }\\
R = \frac{{{v^2}}}{{a \bot }} = \frac{{{{10}^2}}}{{\left( {10\,\cos \,{{30}^ \circ }} \right)}}
\end{array}$
$\begin{array}{l}
\,\, = \frac{{10}}{{\sqrt 3 }} \times 2\, = \frac{{20}}{{\sqrt 3 }}m\\
\frac{{{5^2} + {{\left( {10 – 5\sqrt 3 } \right)}^2}}}{{10\,\cos \theta }} = \frac{{200 – 100\sqrt 3 }}{{10 \times 0.965}} = 2.8m
\end{array}$
