A) (a)\[{{\theta }_{0}}={{\cos }^{-1}}\left( \frac{1}{\sqrt{5}} \right)\]and\[{{\text{v}}_{0}}=\frac{5}{3}m{{s}^{-1}}\]
B) (b)\[{{\theta }_{0}}={{\sin }^{-1}}\left( \frac{1}{\sqrt{5}} \right)\]and\[{{\text{v}}_{0}}=\frac{5}{3}m{{s}^{-1}}\]
C) (c)\[{{\theta }_{0}}={{\sin }^{-1}}\left( \frac{2}{\sqrt{5}} \right)\]and\[{{\text{v}}_{0}}=\frac{3}{5}m{{s}^{-1}}\]
D) (d)\[{{\theta }_{0}}={{\cos }^{-1}}\left( \frac{2}{\sqrt{5}} \right)\]and\[{{\text{v}}_{0}}=\frac{3}{5}m{{s}^{-1}}\]
Correct Answer: A
Solution :
| [a] Equation of trajectory is given as |
| \[y=2x9{{x}^{2}}\] ........ (1) |
| Comparing with equation : |
| \[y=x\tan \theta -\frac{g}{2{{u}^{2}}{{\cos }^{2}}\theta }.{{x}^{2}}\] ........ (2) |
| We get;\[\tan \theta =2\] |
| \[\therefore \] \[\] |
| Also,\[\frac{g}{2{{u}^{2}}{{\cos }^{2}}\theta }=9\] |
| \[\Rightarrow \frac{10}{2\times 9\times {{\left( \frac{1}{\sqrt{5}} \right)}^{2}}}={{u}^{2}}\]\[\Rightarrow {{u}^{2}}=\frac{25}{9}\] |
| \[\Rightarrow \] |
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