A water jet from a function reaches it maximum height of 4 m at a distance 0.5 m from the vertical passing through the point `O` of water outlet. The height of the jet above the horizontal `O X` at a distance of 0.75 m from the point `O` is 5 m (b) 6 m (c) 3 m (d) 7 m

Explanation

The path of a water jet is a parabola.
Let the eqation of this parabolic path be
` y = ax^(2) + bx + c.` ….(i)
Let AB be the maximum height reached.
The path is symmetrical about AB.
Let the jet strike the x-axos at E.
Then, AE = OA = 0.5 m
` rArr OE = 2(OA) = ( 2xx 0.5) m = 1 m.`
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, 1/4 a + 1/2 b = 4 and a+b = 0`
` rArr c = 0, a + 2b = 16 and a + b = 0`
` rArr a =- 16, b = 16 and c = 0.`
`:. ` the equation of the parabola is `y =- 16 x^(2) + 16x,` ....(ii)
Let P be a point on OE such that OP = ` 0.75` m.
Draw `PD bot 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 xx (3/4)^(2) +(16 xx 3/4) rArr h = (-9+12) = 3.`
`:. ` the required height, PD = h m = 3 m.