A) \[R/g\]
B) \[R/4g\]
C) \[2R/g\]
D) \[R/2g\]
Correct Answer: C
Solution :
[c] Range will be same for time \[{{t}_{1}}\And {{t}_{2}},\]so angles of projection will be \['\theta '\And '{{90}^{o}}-\theta '\] |
\[{{t}_{1}}=\frac{2u\sin \theta }{g}{{t}_{2}}=\frac{2u\sin ({{90}^{o}}-\theta )}{g}\] |
and\[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] |
\[{{t}_{1}}{{t}_{2}}=\frac{4{{u}^{2}}\sin \theta \cos \theta }{{{g}^{2}}}=\frac{2}{g}\left[ \frac{2{{u}^{2}}\sin \theta \cos \theta }{g} \right]\]\[=\frac{2R}{g}\] |
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