Question

1. beside a deep river are some cliffs, some of which are over 17 metres tall. a diver climbs to the top of one of the cliffs, and dives into the river below. after hitting the water the diver stays submerged for several seconds before returning to the surface. the vertical position of the diver, with respect to the surface of the water, after t seconds from the time she leaped off the cliff, is modeled by the quadratic function ( h(t)=t^{2}-10t + 17.16 ). at time t, if ( h(t) ) is a positive value, then the diver is ( h(t) ) metres above the surface of the water; and if ( h(t) ) is a negative value, then the diver is ( h(t) ) metres below the surface. how deep under the water is the diver 3 seconds before she reaches the surface?

Explanation:

Step1: Find the time when the diver reaches the surface

When the diver reaches the surface, \(h(t) = 0\). So we solve the quadratic equation \(t^{2}-10t + 17.16=0\).
Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 1\), \(b=-10\), \(c = 17.16\)).
\(\Delta=b^{2}-4ac=(-10)^{2}-4\times1\times17.16=100 - 68.64 = 31.36\)
\(t=\frac{10\pm\sqrt{31.36}}{2}=\frac{10\pm5.6}{2}\)
We get \(t_1=\frac{10 + 5.6}{2}=7.8\) and \(t_2=\frac{10-5.6}{2}=2.2\)

Step2: Calculate the time 3 seconds before reaching the surface

If the diver reaches the surface at \(t = 7.8\) seconds, then the time 3 seconds before is \(t=7.8-3 = 4.8\) seconds.

Step3: Calculate \(h(4.8)\)

Substitute \(t = 4.8\) into \(h(t)=t^{2}-10t + 17.16\)
\(h(4.8)=(4.8)^{2}-10\times4.8+17.16\)
\(h(4.8)=23.04-48 + 17.16\)
\(h(4.8)=-7.8\)

Answer:

The diver is \(7.8\) metres under the water.