Question

quadratic formula with contextual situations

1. during a fireworks show, a firework is launched from a platform 6 m above the ground with an initial upward velocity. its height after t seconds is modeled by ( h(t)=-5 t^{2}+20 t + 6 )

how many seconds pass before the firework hits the ground?

1. a dolphin leaps out of the water, and its height above the surface (in meters) is modeled by ( h(x)=-x^{2}+4 x + 2 )

how far does the dolphin travel horizontally from where it jumped out of the water to where it re-enters?

1. caleb hits a golf ball off a tee that is 1 meter above the ground. the path of the golf ball is modeled by ( g(x)=-4.9 x^{2}+14.7 x + 1 )

how many seconds after he hits the ball does it land back on the ground?

1. a diver is already underwater and swims upward toward the surface. their vertical position relativ the water surface (in meters) is modeled by ( d(x)=3 t^{2}-9 t - 10 ), where ( d(t)=0 ) represents the of the water.

how many seconds does it take the diver

Explanation:

Step1: Set the height function equal to 0

When the firework hits the ground, \(h(t)=0\). So we have the equation \(-5t^{2}+20t + 6=0\). For a quadratic equation \(ax^{2}+bx + c = 0\) (here \(a=-5\), \(b = 20\), \(c = 6\)), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step2: Calculate the discriminant

First, calculate the discriminant \(\Delta=b^{2}-4ac=(20)^{2}-4\times(-5)\times6=400 + 120=520\).

Step3: Substitute into the quadratic formula

\(t=\frac{-20\pm\sqrt{520}}{2\times(-5)}=\frac{-20\pm2\sqrt{130}}{-10}=\frac{-20}{-10}\pm\frac{2\sqrt{130}}{-10}=2\mp\frac{\sqrt{130}}{5}\).
Since time \(t>0\), we take \(t = 2+\frac{\sqrt{130}}{5}\approx2 + 2.28=4.28\) (rounded to two decimal places).

Answer:

\(t\approx4.28\) seconds.