A) 20\[\sqrt{2}\] m
B) 10 m
C) 10\[\sqrt{2}\]m
D) 20 m
Correct Answer: D
Solution :
[d] \[\text{R}\,\,\text{=}\,\,\frac{{{\text{u}}^{2}}{{\sin }^{2}}\theta }{g},\text{ H}\,\,\text{=}\,\,\frac{{{\text{u}}^{2}}{{\sin }^{2}}\theta }{2g}\] \[{{\text{H}}_{\max }}\text{ at 2}\theta \,\,\text{=}\,\,\text{90}{}^\circ \text{ So, }{{\text{H}}_{\max }}=\frac{{{\text{u}}^{2}}}{\text{2g}}\] \[\frac{{{\text{u}}^{2}}}{\text{2g}}=10\Rightarrow {{\text{u}}^{2}}=10g\times 2\] \[R=\frac{{{\text{u}}^{2}}\sin 2\theta }{g}\Rightarrow {{R}_{\max }}=\frac{{{\text{u}}^{2}}}{\text{2g}}=20\text{ meter}\text{.}\]
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