A) \[R=4\sqrt{H{{H}_{1}}}\]
B) \[R=\sqrt{H{{H}_{1}}}\]
C) \[R=H{{H}_{1}}\]
D) None of these
Correct Answer: A
Solution :
Range of projectile \[R=\frac{2{{u}^{2}}\sin \theta \cos \theta }{g}\] ... (i) Height \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] ... (ii) \[{{H}_{1}}=\frac{{{u}^{2}}{{\sin }^{2}}({{90}^{o}}-\theta )}{2g}=\frac{{{u}^{2}}{{\cos }^{2}}\theta }{2g}\] ?. (iii) Then, \[H{{H}_{1}}=\frac{{{u}^{2}}{{\sin }^{2}}\theta \,{{u}^{2}}{{\cos }^{2}}\theta \times 4}{2g\,\,.\,\,2g}\] ... (iv) From Eq. (i), we get \[{{R}^{2}}=\frac{4{{u}^{2}}{{\sin }^{2}}\theta \,{{u}^{2}}{{\cos }^{2}}\theta \times 4}{2g\times 2g}\] \[R=\sqrt{16\,H{{H}_{1}}}\] [from Eq. (iv)] \[=4\sqrt{H{{H}_{1}}}\]
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