Q. A ball projected from ground at an angle of $45^\circ$ 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 wall is:
Solution:
As ball is projected at an angle $45^°$ to the horizontal therefore Range $= 4H$
or $10=4H \Rightarrow H=\frac{10}{4}=2.5\,m$
$\left(\because Range=4m + 6m = 10m\right)$
Maximum height, $H= \frac{u^{2}\,sin^{2}\,\theta}{2g}$
$\therefore u^{2}=\frac{H\times2g}{sin^{2}\,\theta}=\frac{2.5\times2\times10}{\left(\frac{1}{\sqrt{2}}\right)^{2}}=100$
or, $u=\sqrt{100}=10\,ms^{-1}$
Height of wall $PA$
$=OA \,tan\,\theta-\frac{1}{2} \frac{g\left(OA\right)^{2}}{u^{2}\,cos^{2}\theta}$
$=4-\frac{1}{2}\times\frac{10\times16}{10\times10\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}}=2.4\,m$
