A particle is projected from a point \(O\) on a horizontal plane with speed \(V\) and at an angle of elevation \(\alpha\). The vertical plane in which the motion takes place is perpendicular to two vertical walls, both of height \(h\), at distances \(a\) and \(b\) from \(O\). Given that the particle just passes over the walls, find \(\tan\alpha\) in terms of \(a\), \(b\) and \(h\) and show that \[ \frac{2V^2} g = \frac {ab} h +\frac{ (a+b)^2 h}{ab} \;. \] The heights of the walls are now increased by the same small positive amount \(\delta h\,\). A second particle is projected so that it just passes over both walls, and the new angle and speed of projection are \(\alpha +\delta \alpha \) and \(V+\delta V\), respectively. Show that \[ \sec^2 \alpha \, \delta \alpha \approx \frac {a+b}{ab}\,\delta h \;, \] and deduce that \(\delta \alpha >0\,\). Show also that \(\delta V\) is positive if \(h> ab/(a+b)\) and negative if \(h

A particle is projected from a point $O$ on a horizontal plane
with speed $V$ and at an angle
of elevation $\alpha$. The vertical plane in which the motion takes place
is perpendicular to two vertical walls, both of height $h$, at distances
$a$ and $b$ from $O$. Given that the particle just passes over the
walls, find $\tan\alpha$ in terms of $a$, $b$ and $h$ and
show that
\[
\frac{2V^2} g = \frac {ab} h +\frac{ (a+b)^2 h}{ab} \;.
\]
The heights of the walls  are now increased  by the same  small positive
amount $\delta h\,$. 
A  second particle is projected so that it just passes over
both walls,  and  the new angle and speed of projection 
are  $\alpha +\delta \alpha $ and $V+\delta V$, respectively.
Show that 
\[
\sec^2 \alpha \, \delta \alpha  \approx \frac {a+b}{ab}\,\delta h \;,
\]
and deduce that $\delta \alpha >0\,$. Show also that 
$\delta V$ is positive if $h> ab/(a+b)$ and negative if $h<ab/(a+b)\,$.