A) \[\sqrt{{{\operatorname{h}}_{1}}{{h}_{2}}}\]
B) \[\frac{{{h}_{1}}+{{h}_{2}}}{2}\]
C) \[4\sqrt{{{\operatorname{h}}_{1}}{{h}_{2}}}\]
D) \[\frac{2\sqrt{{{h}_{1}}{{h}_{2}}}}{{{h}_{1}}+{{h}_{2}}}\]
Correct Answer: C
Solution :
For the same range, of projectile fired with velocity u, there are two angle of projection, 9 and 90-9 with horizontal. Corresponding height \[{{\operatorname{h}}_{1}}=\frac{{{u}^{2}}si{{n}^{2}}\theta }{2\operatorname{g}}\] \[{{\operatorname{h}}_{2}}=\frac{{{u}^{2}}si{{n}^{2}}\left( 90-\theta \right)}{2\operatorname{g}}=\frac{{{\operatorname{u}}^{2}}co{{s}^{2}}\theta }{2\operatorname{g}}\] \[{{\operatorname{h}}_{1}}{{\operatorname{h}}_{2}}=\frac{{{u}^{2}}si{{n}^{2}}\theta \,{{\cos }^{2}}\theta }{4\operatorname{g}}\] multiply and divide RHS by 4 \[{{h}_{1}}{{\operatorname{h}}_{2}}=\frac{4\,\,{{u}^{4}}si{{n}^{2}}\theta \,{{\cos }^{2}}\theta }{16\operatorname{g}}\] \[16{{h}_{1}}{{\operatorname{h}}_{2}}=\frac{\left( 2{{u}^{2}}sin\theta \,\cos \theta \right)}{{{\operatorname{g}}^{2}}}\] \[16{{h}_{1}}{{\operatorname{h}}_{2}}={{\left[ \frac{{{u}^{2}}sin2\theta }{\operatorname{g}} \right]}^{2}}\] \[R=\frac{{{u}^{2}}sin2\theta }{\operatorname{g}}\] \[16{{h}_{1}}{{h}_{2}}={{R}^{2}}\] \[R=4\sqrt{{{\operatorname{h}}_{1}}{{h}_{2}}}\]
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