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 ONLINE 22-04-2013]

A) 4.4 m

B) 2.4 m

C)             3.6 m

D) 1.6 m

Correct Answer: B

Solution :

[b]

As ball is projected at an angle \[{{45}^{o}}\]to the horizontal therefore Range = 4H

or         \[10=4H\Rightarrow H=\frac{10}{4}=2.5\,m\]

(\[\because \]\[Rang=4m+6m=10m\])

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

\[=OA\tan \theta -\frac{1}{2}\frac{g{{(OA)}^{2}}}{{{u}^{2}}{{\cos }^{2}}\theta }\]

\[=4-\frac{1}{2}\times \frac{10\times 16}{10\times 10\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}}=2.4\,\,m\]