A) \[\frac{{{R}^{2}}}{8h}-2h\]
B) \[\frac{{{R}^{2}}}{8h}-2gh\]
C) \[\frac{{{R}^{2}}}{8h}+2h\]
D) \[\frac{{{R}^{2}}}{8h}\]
Correct Answer: C
Solution :
[c] \[R=\frac{{{u}^{2}}Sin\,2\theta }{g}\] \[{{R}^{2}}=\frac{{{u}^{2}}Sin\,2\theta \,{{\cos }^{2}}\theta }{{{g}^{2}}}\,\,.......(1)\] \[h-\frac{{{u}^{2}}\,\sin \,\theta }{2g}\,\,\,......(2)\,\] \[(1)\div (2)\] \[\frac{{{R}^{2}}}{8h}=\frac{{{u}^{2}}{{\cos }^{2}}\theta }{g}...........(3)\] \[2h=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{g}\] \[\frac{{{R}^{2}}}{8h}+2h=\frac{{{u}^{2}}}{g}\left[ {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right]=\frac{{{u}^{2}}}{g}={{R}_{\max }}\]
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