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 \[V=10\,m/s\] |
\[\therefore \] \[r=100\sqrt{10}\,m.\] |
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