A) \[15\left( \sqrt{3}-1 \right)\text{s}\]
B) \[15\left( \sqrt{3}+1 \right)\text{s}\]
C) \[7.5\left( \sqrt{3}-1 \right)\text{s}\]
D) \[7.5\left( \sqrt{3}+1 \right)\text{s}\]
Correct Answer: C
Solution :
| [c] At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. |
| Then \[\text{u cos 60 }\!\!{}^\circ\!\!\text{ = v cos 4}5{}^\circ .\] |
| \[\Rightarrow 150\times \frac{1}{2}=\text{v}\times \frac{1}{\sqrt{2}}\text{ or }\frac{150}{\sqrt{2}}\text{ m/s}\] |
| \[\text{Initially: }{{\text{u}}_{y}}=u\sin 60{}^\circ =\frac{150\sqrt{3}}{2}\text{ m/s}\] |
| \[\text{Finally: }{{\text{v}}_{y}}=v\sin 45{}^\circ =\frac{150}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{150}{2}\text{ m/s}\] |
| \[\text{But }{{\text{v}}_{y}}={{u}_{y}}+{{a}_{y}}t\text{ or }\frac{150}{2}=\frac{150\sqrt{3}}{2}-10t\] |
| \[10\,t=\frac{150}{2}\left( \sqrt{3}-1 \right)\text{ or t}\,\text{= 7}\text{.5}\left( \sqrt{3}-1 \right)s\] |
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