A ball projected from ground at an angle of \(45^o\) just clears a wall in front. If point of projection is \(4\,m\) 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 walI is  ........ \( m\)

\(\begin{array}{l}
As\,ball\,is\,projected\,at\,an\,angle\,{45^{ \circ \,}}to\,the\\
horizontal\,therefore\,Rang = 4H\\
or\,\,\,\,10 = 4H \Rightarrow H\frac{{10}}{4} = 2.5m\\
\left( {Range\, = 4m\, + 6\,m = 10m} \right)\\
Maximum\,height,\,H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\\
\therefore \,{u^2} = \frac{{H \times 2g}}{{{{\sin }^2}\theta }} = \frac{{2.5 \times 2 \times 10}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}} = 100
\end{array}\)

\(\begin{array}{l}
or,\,u = \sqrt {100}  = 10\,m{s^{ - 1}}\\
Height\,of\,wall\,PA\\
 = OA\,\tan \,\theta  - \frac{{1\,g{{\left( {OA} \right)}^2}}}{{2\,{u^2}\,{{\cos }^2}\theta }}\\
 = 4 = \frac{1}{2} \times \frac{{10 \times 16}}{{10 \times 10 \times \frac{1}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }}}} = 2.4m
\end{array}\)