A) 1
B) \[\frac{1}{2}\]
C) \[ta{{n}^{2}}\theta \]
D) None of these
Correct Answer: A
Solution :
In case of projectile, The horizontal range \[R=\frac{{{u}^{2}}}{g}\sin \,2\theta \] In first case \[{{R}_{1}}=\frac{{{u}^{2}}\sin \,2\theta }{g}\] In second case \[{{\operatorname{R}}_{2}}=\frac{{{u}^{2}}\,sin 2\left( 90{}^\circ - \theta \right)}{g}\] \[=\,\,\,\frac{{{u}^{2}}\,sin\left( 180{}^\circ - 2\,\theta \right)}{g}\,\,=\,\,\frac{{{u}^{2}}\sin \,2\,\theta }{g}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{R}_{1}}}{{{R}_{2}}}=1\]
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