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