Question

At a water fountain, water attains a maximum height of 4m at horizontal

distance of 0.5 from its origin. If the path of water is a parabola, find the

height of water at a horizontal distance of 0.75 from the point of origin.​

Answer 1

Answer:

From the given information, the parabola is symmetric about y-axis and it is open down word.

Since the vertex is (h, k),

(x - h)2 = -4a (y - k)

Here (h, k) ==> (0, 10)

(x - 0)2 = -4a (y - 10)

x2 = -4a (y - 10) ----(1)

The parabola is passing through the point (15, 0).

152 = -4a (0 - 10)

225/40 = a

a = 45/8

By applying the value of a in (1), we get

x2 = -4 (45/8) (y - 10)

x2 = -(45/2) (y - 10)

By applying x = 6 and y = h, we get

62 = -(45/2) (h - 10)

-36(2/45) = h - 10

(-72/45) + 10 = h

h = -1.6 + 10

h = 8.4 m

Hence the required height is 8.4 m.

Answer 2

Given:

height=4m

horizontal distance from origin=0.5m

To find:

height of water at a horizontal distance of 0.75m

Solution:

The general equation of a parabola that is open downward is:

$(x-h)^2=-4a(y-k)$  -(1)

The vertex of the parabola is at $(\frac{1}{2}, 4)$

Putting these values in (1):

$(x-\frac{1}{2})^2=-4a(y-4)$  -(2)

Now, the parabola also passes through the origin O(0, 0), so,

$(0-\frac{1}{2})^2=-4a(0-4)$

⇒$\frac{1}{4} =16a$

⇒$a=\frac{1}{64}$

Putting the value of a in (1), we get,

$(x-\frac{1}{2} )^2=-4$×$\frac{1}{64}$×$(y-4)$

Now, we put the value (0.75, h), where h is the required value of height:

$(0.75-0.5)^2=-\frac{1}{16}(h-4)$

⇒$(0.25)^2=-\frac{1}{16}(h-4)$

⇒$h-4=-0.0625$×$16$

⇒$h=4-1$

⇒$h=3$

Hence, height of water at a horizontal distance of 0.75m is 3m.