A) \[15\,\,s\]
B) \[10.98\,\,s\]
C) \[5.49\,\,s\]
D) \[2.745\,\,s\]
Correct Answer: C
Solution :
Key Idea: At the two points of the trajectory during projection, the horizontal component of the velocity is the same. Horizontal component of velocity at angle\[{{60}^{o}}\] \[=\]Horizontal component of velocity at\[{{45}^{o}}\] \[i.e.,\] \[u\cos {{60}^{o}}=v\sin {{45}^{o}}\] or \[147\times \frac{1}{2}=v\times \frac{1}{\sqrt{2}}\] or \[v=\frac{147}{\sqrt{2}}m/s\] Vertical component of\[u=u\sin {{60}^{o}}\] \[=\frac{147\sqrt{3}}{2}m\] Vertical component of\[v=v\sin {{45}^{o}}\] \[=\frac{147}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\] \[=\frac{147}{2}m\] but \[{{v}_{y}}={{u}_{y}}+at\] \[\therefore \] \[\frac{147}{2}=\frac{147\sqrt{3}}{2}-9.8t\] or \[9.8\,\,t=\frac{147}{2}(\sqrt{3}-1)\] \[\therefore \] \[t=5.49\,\,s\]
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