Question

1. a small town has installed two fountain jets in the town square.

after the first jet is fired, the water reaches a maximum height of 3.6 m in 2.4 seconds.
the second jet is fired 3.6 seconds after the first jet.
both jets then are at a height of 2.0 m at the same time.
the second jet of water hits the ground 8.0 seconds after the first jet is fired.
two second - degree polynomial functions represent the heights of the jets in relation to the
time elapsed from the moment the first jet is fired. the two functions are represented by
the graph below.
where x: time elapsed, in seconds, from the moment the first jet is fired
f(x): the height of the water of the first jet, in metres
g(x): the height of the water of the second jet, in metres
what is the maximum height reached by the water fired by the second jet?

Explanation:

Step1: First jet's function

The first jet \( f(x) \) is a quadratic function. Since it has a maximum at \( x = 2.4 \) and passes through \( (0,0) \), its equation is \( f(x)=a(x - 2.4)^{2}+3.6 \). Substituting \( (0,0) \) gives \( 0=a(0 - 2.4)^{2}+3.6 \), so \( a=-\frac{3.6}{2.4^{2}}=-\frac{3.6}{5.76}=-\frac{5}{8} \). Thus \( f(x)=-\frac{5}{8}(x - 2.4)^{2}+3.6 \).

Step2: Solve for intersection time

We know \( f(x)=g(x) = 2.0 \). For \( f(x)=2.0 \), we have \( 2.0=-\frac{5}{8}(x - 2.4)^{2}+3.6 \). Then \( \frac{5}{8}(x - 2.4)^{2}=1.6 \), \( (x - 2.4)^{2}=\frac{1.6\times8}{5}=2.56 \), \( x - 2.4=\pm1.6 \). So \( x = 4 \) or \( x = 0.8 \). Since the second jet starts at \( x = 3.6 \), we take \( x = 4 \).

Step3: Second jet's function

The second jet \( g(x) \) is a quadratic function. It passes through \( (3.6,0) \) and \( (8.0,0) \), so its equation is \( g(x)=b(x - 3.6)(x - 8.0) \). Substituting \( (4,2.0) \) gives \( 2.0=b(4 - 3.6)(4 - 8.0) \), \( 2.0=b\times0.4\times(- 4) \), \( b=-\frac{2.0}{1.6}=-\frac{5}{4} \). Then \( g(x)=-\frac{5}{4}(x - 3.6)(x - 8.0)=-\frac{5}{4}(x^{2}-11.6x + 28.8) \).

Step4: Find maximum of second jet's function

The \( x \) - coordinate of the vertex of \( y = ax^{2}+bx + c \) is \( x=-\frac{b}{2a} \). For \( g(x)=-\frac{5}{4}x^{2}+\frac{5\times11.6}{4}x-\frac{5\times28.8}{4} \), \( a =-\frac{5}{4} \), \( b=\frac{5\times11.6}{4} \). \( x=\frac{11.6}{2}=5.8 \). Then \( g(5.8)=-\frac{5}{4}(5.8 - 3.6)(5.8 - 8.0)=-\frac{5}{4}\times2.2\times(-2.2)=6.05 \)

Answer:

\( 6.05\ m \)