Since, the height and width of the arc from the centre is 2m and 8m respectively, it is clear that the length of the major axis is 8m, while the length of the semi- minor axis is 2m.

The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis.

The equation of the semi – ellipse will be of the from \( \dfrac{ x^2}{16} + \dfrac{y^2}{4}= 1\), y ≥ 0 … (1

Let A be a point on the major axis such that AB = 1.5m.

Now draw AC ⊥ OB.

OA = (4 – 1.5)m = 2.5m

The x – coordinate of point C is 2.5

On substituting the value of x with 2.5 in equation (1), we get,

\( \dfrac{ (2.5)^2}{16} + \dfrac{y^2}{4}= 1\)

\( \dfrac{ 6.25}{16} + \dfrac{y^2}{4}= 1\)

y^{2} = 4 (1 – \( \dfrac{ 6.25}{16}\))

= \( 4( \dfrac{ 9.75}{16})\)

= 2.4375

y = 1.56 (approx.)

So, AC = 1.56m

Hence, the height of the arch at a point 1.5m from one end is approximately 1.56m.

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