A) 15s
B) 10.98s
C) 5.49s
D) 2.745s
Correct Answer: C
Solution :
Initial velocity of projectile,\[u=147\text{ }m/s\] At the two points of the trajectory during projection, the horizontal component of the velocity is the same. \[u\cos {{60}^{o}}=v\cos {{45}^{o}}\] \[147\times \frac{1}{2}=v\times \frac{1}{\sqrt{2}}\] \[v=\frac{147}{\sqrt{2}}m/s\] Vertical component of \[u=u\sin {{60}^{o}}\] \[=\frac{147\sqrt{3}}{2}m\] Vertical component of\[v=v\sin {{45}^{o}}\] \[=\frac{147}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\] \[=\frac{147}{2}m\] \[{{v}_{y}}={{u}_{y}}-gt\] \[\frac{147}{2}=\frac{147\sqrt{3}}{2}-9.8t\] \[9.8t=\frac{147}{2}(\sqrt{3}-1)\] \[t=\frac{147}{2\times 9.8}(\sqrt{3}-1)\] \[=5.49s\]You need to login to perform this action.
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