A) 2.5 m
B) 5 m
C) 10 m
D) 20 m
Correct Answer: B
Solution :
[b] At a time t when velocity vector become mutually perpendicular \[v\text{ }cos\text{ }45{}^\circ =5\] horizontal component |
\[v=\frac{5}{cos\text{ }45{}^\circ }=5\sqrt{2}\text{ m/s }\] |
Vertically, |
\[v\,\,\sin 45{}^\circ =gt\] |
\[\Rightarrow t=\frac{v\,\,\sin 45{}^\circ }{g}=\frac{5\sqrt{2}\times 1/\sqrt{2}}{10}=\frac{5}{10}=\frac{1}{2}\] |
So, \[OA=OB=v\cos 45{}^\circ \times t=5\times \frac{1}{2}=2.5\] |
\[\Rightarrow \,\,\,AB=2.5\times 2=5\text{ m}\] |
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