Q25.

Question

Della’s parents are throwing a Sweat 16 party for her. At 10:00, a ball will slide 25 feet down a pole and light up. A function that models the drop is $h = - t^{2} + 5 t + 25,$ where $h$ is in feet of the ball after $t$ seconds. How many seconds will it take for the ball to reach the bottom of the pole?

Step-by-Step Solution

Verified

Answer

After $8 .090169944$ seconds the ball will reach the bottom of the pole.

1Step 1. Define the standard form of the quadratic equation.

A quadratic equation, which is written in the form, $a x^{2} + b x + c = 0$ , where, $a \neq 0$ is called the standard form of the quadratic equation.

2Step 2. Define the quadratic formula.

For the quadratic equation $a x^{2} + b x + c = 0,$ where, $a \neq 0$ is given by

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

3Step 3. Calculate the time taken by the ball to reach the bottom of the pole.

The height of the ball in feet after $t$ seconds is given by

$h = - t^{2} + 5 t + 25$

When the ball will reach the bottom of the pole, then at that time, $h = 0$ .

So, substitute $h = 0$ in $h = - t^{2} + 5 t + 25$ .

$0 = - t^{2} + 5 t + 25 - t^{2} + 5 t + 25 = 0$

Compare the quadratic equation $- t^{2} + 5 t + 25 = 0$ with the quadratic equation $a t^{2} + b t + c = 0$ .

$a = - 1, b = 5, c = 25$

Substitute, $a = - 1, b = 5$ and $c = 25$ in $t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ .

$t = \frac{- 5 \pm \sqrt{5^{2} - 4 (- 1) (25)}}{2 (- 1)} t = \frac{- 5 \pm \sqrt{25 + 100}}{- 2} t = \frac{- 5 \pm \sqrt{125}}{- 2} t = \frac{- 5 \pm 5 \sqrt{5}}{- 2} t = \frac{- 5 - 5 \sqrt{5}}{- 2}, \frac{- 5 + 5 \sqrt{5}}{- 2} t = \frac{5 + 5 \sqrt{5}}{2}, \frac{5 - 5 \sqrt{5}}{2}$