A) \[R=\frac{{{u}^{2}}\,\sin \,\,2\theta }{g}\]
B) \[R=\frac{{{u}^{2}}\,{{\sin }^{2}}\theta }{2g}\]
C) \[R=\frac{{{u}^{2}}}{g}\]
D) \[R={{u}^{2}}\,\,\sin \,\,\theta \]
Correct Answer: C
Solution :
\[R=\frac{{{u}^{2}}\,\sin \,2\theta }{g},\] At \[\theta ={{45}^{o}},\,R=\max .\] \[\therefore \] \[{{R}_{\max }}=\frac{{{u}^{2}}}{g}\]
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