A) \[{{\text{t}}_{\text{1}}}{{\text{t}}_{\text{2}}}\propto \text{R}\]
B) \[{{\text{t}}_{\text{1}}}{{\text{t}}_{\text{2}}}\propto {{\text{R}}^{2}}\]
C) \[{{\text{t}}_{\text{1}}}{{\text{t}}_{\text{2}}}\propto 1/\text{R}\]
D) \[{{\text{t}}_{\text{1}}}{{\text{t}}_{\text{2}}}\propto 1/{{\text{R}}^{2}}\]
Correct Answer: A
Solution :
[a] \[{{\text{t}}_{1}}=\frac{2\,u\,\,\sin \theta }{\text{g}}\text{ and}\] \[{{\text{t}}_{2}}=\frac{2\,u\,\,\sin \left( 90-\theta \right)}{\text{g}}\text{ =}\frac{2\,u\,\,\cos \theta }{\text{g}}\] \[\therefore \,\,\,{{\text{t}}_{1}}{{\text{t}}_{2}}=\frac{4\,{{\text{u}}^{2}}cos\theta \,\,\sin \theta }{{{\text{g}}^{2}}}=\frac{2}{\text{g}}\left[ \frac{{{\text{u}}^{2}}\sin 2\theta }{\text{g}} \right]\]\[=\frac{2}{\text{g}}\text{R,}\] Where R is the range. Hence \[{{\text{t}}_{1}}{{\text{t}}_{2}}\propto \text{R}\]
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