A) \[0{}^\circ \]
B) \[si{{n}^{-1}}\left( \frac{1}{8} \right)\]
C) \[si{{n}^{-1}}\left( \frac{1}{6} \right)\]
D) \[si{{n}^{-1}}\left( \frac{1}{2} \right)\]
Correct Answer: C
Solution :
| [c] \[\text{max heigth, }{{\text{H}}_{\text{A}}}=\frac{\text{u}_{\text{A}}^{2}{{\sin }^{2}}30{}^\circ }{2\text{g}};\] |
| \[{{\text{H}}_{\text{B}}}=\frac{\text{u}_{\text{B}}^{2}{{\sin }^{2}}\theta }{2\text{g}}\] |
| \[\text{As we know, }{{\text{H}}_{\text{A}}}\text{=}\,\,{{\text{H}}_{\text{B}}}\text{ }\] |
| \[\frac{\text{u}_{\text{A}}^{2}{{\sin }^{2}}30{}^\circ }{2\text{g}}=\frac{\text{u}_{\text{B}}^{2}{{\sin }^{2}}\theta }{2\text{g}}\] |
| \[\Rightarrow \frac{{{\sin }^{2}}\theta }{{{\sin }^{2}}30{}^\circ }=\frac{\text{u}_{\text{A}}^{2}}{\text{u}_{\text{B}}^{2}}\] |
| \[{{\sin }^{2}}\theta ={{\left( \frac{{{u}_{A}}}{{{u}_{B}}} \right)}^{2}}{{\sin }^{2}}30{}^\circ \Rightarrow {{\sin }^{2}}\theta =\frac{1}{36}\] |
| \[\sin \theta =\frac{1}{6}\Rightarrow \theta ={{\sin }^{-1}}\left( \frac{1}{6} \right)\] |
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