question_answer

A Ball projected from ground at an angle of \[{{45}^{0}}\] just clears a wall in front. If point of projection is 4m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is:     JEE Main  Online Paper (Held On 22 April 2013)

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 \[=OA\tan \theta -\frac{1}{2}\frac{g{{(OA)}^{2}}}{{{u}^{2}}{{\cos }^{2}}\theta }\] \[=4-\frac{1}{2}\times \frac{10\times 16}{10\times 10\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}}=2.4\,\,m\]