A) \[R=4\sqrt{{{H}_{1}}{{H}_{2}}}\]
B) \[R=\,4({{H}_{1}}-{{H}_{2}})\]
C) \[R=\,4({{H}_{1}}+{{H}_{2}})\]
D) \[R=\,\frac{{{H}^{2}}_{1}}{{{H}^{2}}_{2}}\]
Correct Answer: A
Solution :
[a] \[{{H}_{1}}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}and\,\,\,{{H}_{2}}=\frac{{{u}^{2}}{{\sin }^{2}}(90-\theta )}{2g}=\frac{{{u}^{2}}{{\cos }^{2}}\theta }{2g}\] \[{{H}_{1}}{{H}_{2}}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\times \frac{{{u}^{2}}{{\cos }^{2}}\theta }{2g}\] \[=\frac{{{({{u}^{2}}sin2\theta )}^{2}}}{16{{g}^{2}}}=\frac{{{R}^{2}}}{16}\] \[[\therefore R=4\sqrt{{{H}_{1}}{{H}_{2}}}\]
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