A) 4.4 m
B) 2.4 m
C) 3.6 m
D) 1.6 m
Correct Answer: B
Solution :
As ball is projected at an angle \[{{45}^{o}}\]to the horizontal therefore Range = 4H or \[10=4H\Rightarrow H=\frac{10}{4}=2.5\,m\] (\[Rang=4m+6m=10m\]) Maximum height, \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[\therefore \] \[{{u}^{2}}=\frac{H\times 2g}{{{\sin }^{2}}\theta }=\frac{2.5\times 2\times 10}{{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}}=100\] or, \[u=\sqrt{100}=10m{{s}^{-1}}\] Height of wall PA \[=OA\tan \theta -\frac{1}{2}\frac{g{{(OA)}^{2}}}{{{u}^{2}}{{\cos }^{2}}\theta }\] \[=4-\frac{1}{2}\times \frac{10\times 16}{10\times 10\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}}=2.4\,\,m\]You need to login to perform this action.
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