A) \[{{\sin }^{2}}\theta \]
B) \[{{\cos }^{2}}\theta \]
C) \[{{\tan }^{2}}\theta \]
D) 1
Correct Answer: C
Solution :
In case of projectile, the maximum height \[H=\frac{{{u}^{2}}\,si{{n}^{2}}\,\theta }{2g}\] In first case, \[{{H}_{1}}=\frac{{{u}^{2}}\,si{{n}^{2}}\,\theta }{2g}\] ... (i) In second case, \[{{H}_{2}}=\frac{{{u}^{2}}\,si{{n}^{2}}(90-\,\theta )}{2g}\,\,=\,\,\frac{{{u}^{2}}{{\cos }^{2}}\theta }{2g}\] ? (ii) Now, \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{\frac{{{u}^{2}}{{\sin }^{2}}\,\theta }{2g}}{\frac{{{u}^{2}}{{\sin }^{2}}\,\theta }{2g}}\,\,=\,\,\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }={{\tan }^{2}}\theta \] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{H}_{1}}}{{{H}_{2}}}={{\tan }^{2}}\theta \]
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