A) 336
B) 224
C) 56
D) 34
Correct Answer: D
Solution :
Given : For first stone angle of projection \[{{\theta }_{1}}=\frac{\pi }{3}={{60}^{o}}\] Maximum height reached by it \[{{h}_{1}}=102\,m\] For second stone angle of projection \[{{\theta }_{2}}=\frac{\pi }{2}-{{\theta }_{1}}=\frac{\pi }{2}-\frac{\pi }{3}=\frac{\pi }{6}=\frac{180}{6}={{30}^{o}}\] (As the ranges are same) Maximum height reached by first stone is given by \[{{H}_{1}}=\frac{{{u}^{2}}{{\sin }^{2}}{{\theta }_{1}}}{2g}\] ?(i) Maximum height reached by \[{{H}_{2}}=\frac{{{u}^{2}}{{\sin }^{2}}{{\theta }_{2}}}{2g}\] ?(ii) Now, dividing Eq. (ii) by (i), we get \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{u}^{2}}{{\sin }^{2}}{{\theta }_{1}}/2g}{{{u}^{2}}{{\sin }^{2}}{{\theta }_{2}}/2g}=\frac{{{\sin }^{2}}{{\theta }_{1}}}{{{\sin }^{2}}{{\theta }_{2}}}=\frac{{{\sin }^{2}}{{60}^{o}}}{{{\sin }^{2}}{{30}^{o}}}\] \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}{{{\left( \frac{1}{2} \right)}^{2}}}\] \[{{H}_{2}}=\frac{1}{4}\times \frac{4}{3}\times {{H}_{1}}=\frac{{{H}_{1}}}{3}\] \[{{H}_{2}}=\frac{102}{3}=34m\]
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