A) \[{{t}_{1}}{{t}_{2}}\propto R\]
B) \[{{t}_{1}}{{t}_{2}}\propto {{R}^{2}}\]
C) \[{{t}_{1}}{{t}_{2}}\propto \frac{1}{{{R}^{2}}}\]
D) \[{{t}_{1}}{{t}_{2}}\propto \frac{1}{R}\]
Correct Answer: A
Solution :
\[{{t}_{1}}=\frac{2u\,\sin \alpha }{g}\] \[{{t}_{2}}=\frac{2u\,\sin (90-\alpha )}{g}\] So, \[{{t}_{1}}\times {{t}_{2}}=2\frac{{{u}^{2}}}{{{g}^{2}}}\sin 2\alpha \] or \[{{t}_{1}}\times {{t}_{2}}=\frac{2R}{g}\] \[\left( \because \,\,R=\frac{{{u}^{2}}\sin 2\alpha }{g} \right)\] \[{{t}_{1}}\times {{t}_{2}}\propto R\]
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