A) \[\sqrt{3}-1\]
B) \[1/\sqrt{3}-1\]
C) \[\sqrt{3}+1\]
D) \[1/\sqrt{3}+1\]
Correct Answer: A
Solution :
\[{{x}_{2}}=u\cos t\,\theta t,{{x}_{2}}-{{x}_{1}}=vt\] \[\Rightarrow \] \[{{x}_{1}}+vt=u\cos \theta t\] \[h=x\tan \theta -\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }\] \[\frac{3h}{2}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] Simplify to get, \[{{x}^{2}}-6h\,\cot \theta x+6{{h}^{2}}{{\cot }^{2}}\theta =0\] \[x=\frac{6h\cot \theta \pm \sqrt{12}h\cot \theta }{2}\] \[=3h\cot \theta \pm \sqrt{3h}\cot \,\theta \] \[{{x}_{1}}=3h\cot \,\theta -\sqrt{3h}\cot \theta \] and \[{{x}_{2}}=3h\cot \theta +\sqrt{3h}\cot \theta \] \[v=\frac{{{x}_{2}}-{{x}_{1}}}{t}=\frac{2\sqrt{3h}\cot \theta }{t}\] \[u\cos \theta =\frac{{{x}_{2}}}{t}=\frac{(3+\sqrt{3})h\cot \theta }{t}\] \[\frac{v}{u\cos \theta }=\frac{2\sqrt{3}}{3+\sqrt{3}}=\frac{2\sqrt{3}(3-\sqrt{3})}{9-3}=\sqrt{3}-1\]You need to login to perform this action.
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