Since it is given that the arch is $8 \mathrm{~m}$ wide and $2 \mathrm{~m}$ high at centre in the form of a semi-ellipse
$\therefore$ Arch is part of ellipse whose semi-major and semiminor axis are $4 \mathrm{~m}$ and $2 \mathrm{~m}$ respectively
i.e. $a=4, b=2$
$\therefore \quad$ its eqn is $\frac{x^2}{16}+\frac{y^2}{4}=1 \Rightarrow x^2+4 y^2=16$


A point $1.5$ from one end of arch is $(4-1.5)=2.5 \mathrm{~m}$ from origin, height of arch at this pt. is given by the value of it obtained by putting $\quad x=2.5$
$\begin{aligned}
&\therefore \quad(2.5)^2+4 y^2=16 \\
&\quad 4 y^2=16-6.25=9.75
\end{aligned}$
$y=\frac{\sqrt{9.75}}{2}=\frac{3.12}{2}=1.56 \mathrm{~m} \text { (approx) }$
$\therefore \quad$ Height of the arch is $1.56 \mathrm{~m}$ approx.