question_answer

A stone is thrown horizontally under gravity with a speed of 10m/sec. Find the radius of curvature of it's trajectory at the end of 3 sec after motion began.

A) \[10\sqrt{10\,}\,m\]

B) \[100\sqrt{10\,}\,m\]

C) \[\sqrt{10\,}\,m\]

D) \[100\,\,m\]

Correct Answer: B

Solution :

Method (I)

After 3 sec. \[{{V}_{y}}={{u}_{y}}+gt=-\,30\,m/s\]

and  \[{{V}_{x}}=10\,m/s\]

\[\therefore \] \[{{V}^{2}}=V_{x}^{2}+V_{y}^{2}\]\[\Rightarrow \]\[V=10\sqrt{10}\,m/s\] Now, \[\tan \alpha =\frac{{{V}_{x}}}{{{V}_{y}}}=\frac{1}{3}\]\[\Rightarrow \]\[\sin \alpha =\frac{1}{\sqrt{10}}\]

Radius of curvature \[r=\frac{V_{\bot }^{2}}{g\sin \alpha }\]

\[r=100\sqrt{10}\,m\]

Method (II)

Let horizontal and vertical position of point p be x & y respectively

\[\therefore \]      \[x=Vt\] and  \[y=\frac{1}{2}\,\,g{{t}^{2}}\]

\[\therefore \]      equation of trajectory \[y=\frac{g{{x}^{2}}}{2{{V}^{2}}}\]

\[\therefore \]      \[\frac{dy}{dx}=\frac{gx}{{{V}^{2}}}\] and \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{g}{{{V}^{2}}}\]

Radius of curvature \[r=\frac{{{\left\{ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right\}}^{3/2}}}{\frac{{{d}^{2}}y}{dx}}={{\frac{\left( 1+\frac{{{g}^{2}}{{x}^{2}}}{{{V}^{4}}} \right)}{g/{{v}^{2}}}}^{3/2}}\]Now after 3 s \[x=Vt=30\,m\] and \[V=10\,m/s\]

\[\therefore \]      \[r=100\sqrt{10}\,m.\]