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\[60{}^\circ \] = Horizontal component of velocity at\[45{}^\circ \] i.e., \[u\cos 60{}^\circ =v\,\sin 45{}^\circ \] or \[147\times \frac{1}{2}=v\times \frac{1}{\sqrt{2}}\] or \[v=\frac{147}{\sqrt{2}}m\text{/}s\] Vertical component of \[u=u\sin 60{}^\circ \] \[=\frac{147\sqrt{3}}{2}m\] Vertical component of\[v=v\sin 45{}^\circ \] \[=\frac{147}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{147}{2}m\] but \[{{v}_{y}}={{u}_{y}}+at\] \[\therefore \] \[\frac{147}{2}=\frac{147\sqrt{3}}{2}-9.8\,t\] or \[9.8\,t=\frac{147}{2}(\sqrt{3}-1)\] \[\therefore \] \[t=5.49\,s\]
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