Question

A water jet coming out of the small opening O of a fountain reaches its maximum height of 4 metres at a distance of 0.5 metre from the vertical. Find the height of the jet above the horizontal OX at a distance of 0.75 metre from the point O.

Answer 1

The path of a water jet is a parabola.

Let the equation of this parabolic path be $y=a{x}^2+bx+c....\left(i\right)$

Let AB be the maximum height reached.

The path is symmetrical about AB.

Let the jet strike the x-axis at E.

Then, $AE=OA=0.5m$

$\Rightarrow OE=2\left(OA\right)=(2\times 0.5)m=1m.$

Clearly AB = 4 m.

Thus, the points O, B and E are

O(0, 0) B(0.5, 4) and E(1, 0).

Since these points lie on (i), we have

$c=0,\frac{1}{4}a+\frac{1}{2}b=4$ and $a+b=0$

$\Rightarrow c=0,a+2b=16$ and $a+b=0$

$\Rightarrow a=-16,b=16$ and $c=0.$

$\therefore$ the equation of the parabola is $y=-16x^2+16x....\left(ii\right)$

Let P be a point on OE such that OP = 0.75 M.

Draw $PD\perp OX$, meeting the parabola at D.

Let PD = h metres.

Then, the coordinates of D are (0.75, h).

Since D(0.75, h) lies on (ii), we have

$h=-16\times {\left(\frac{3}{4}\right)}^2+(16\times \frac{3}{4})\Rightarrow h=(-9+12)=3.$

$\therefore$ the required height, PD = h m = 3 m.