A) 200 m/s
B) 100 m/s
C) 400 m/s
D) 300 m/s
Correct Answer: C
Solution :
\[R={{\mu }_{n}}\times T\] \[=u\cos \theta \times \frac{2\mu \sin \theta }{g}\] \[={{u}^{2}}\left( \frac{2\sin \theta .\cos \theta }{g} \right)\] \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] For \[{{R}_{\max }}=\sin 2\theta =1\] \[{{R}_{\max }}=\frac{{{u}^{2}}}{g},u=\sqrt{{{R}_{\max }}g}\] \[u=\sqrt{16\times {{10}^{3}}\times 10}\] \[u=400\,m/s\]
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