A) 336
B) 224
C) 56
D) 34
Correct Answer: D
Solution :
We know that if two stones have same horizontal range, then this implies that both are projected at\[\theta \]and\[{{90}^{o}}-\theta \]. Here, \[\theta =\frac{\pi }{3}={{60}^{o}}\] \[{{90}^{o}}-\theta ={{90}^{o}}-{{60}^{o}}={{30}^{o}}\] For first stone, Max. height\[=102=\frac{{{u}^{2}}{{\sin }^{2}}{{30}^{o}}}{2g}\] For second stone, Max. height,\[h=\frac{{{u}^{2}}{{\sin }^{2}}{{30}^{o}}}{2g}\] \[\therefore \] \[\frac{h}{102}=\frac{{{\sin }^{2}}{{30}^{o}}}{{{\sin }^{2}}{{60}^{o}}}=\frac{{{(1/2)}^{2}}}{{{(\sqrt{3}/2)}^{2}}}\] or \[h=102\times \frac{1/4}{3/4}=34\,\,m\]
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