A) \[\frac{x\sin {{\theta }_{2}}_{\,}}{{{u}_{1}}}\]
B) \[\frac{x\text{ cos}{{\theta }_{2}}_{\,}}{{{u}_{2}}}\]
C) \[\frac{x\sin {{\theta }_{2}}_{\,}}{{{u}_{1}}\sin \left( {{\theta }_{2}}-{{\theta }_{1}} \right)}\]
D) \[\frac{x\sin {{\theta }_{1}}_{\,}}{{{u}_{2}}\sin \left( {{\theta }_{2}}-{{\theta }_{1}} \right)}\]
Correct Answer: C
Solution :
| [c] \[x+{{u}_{2}}\cos {{\theta }_{2}}t={{u}_{1}}\cos {{\theta }_{1}}t\] |
| \[\therefore \,\,\,\,\,\,\,\,t=\frac{x}{{{u}_{1}}\cos {{\theta }_{1}}-{{u}_{2}}\cos {{\theta }_{2}}}\] ?(i) |
| Also \[{{u}_{1}}\sin {{\theta }_{1}}={{u}_{2}}\sin {{\theta }_{2}}\] ?(ii) |
| After solving above equations, we get |
| \[t=\frac{x\,\sin {{\theta }_{2}}}{u{{ & }_{1}}\sin \left( {{\theta }_{2}}-{{\theta }_{1}} \right)}\] |
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