Question

a ball is thrown from an initial height of 4 feet with an initial upward velocity of 17 ft/s. the balls height \\(h\\) (in feet) after \\(t\\) seconds is given by the following.

\\h = 4 + 17t - 16t^2\\

find all values of \\(t\\) for which the balls height is 8 feet.

round your answer(s) to the nearest hundredth.
(if there is more than one answer, use the \or\ button.)

Explanation:

Set up the equation

We are given the height equation:

$$h = 4 + 17t - 16t^2$$

We want to find the values of \(t\) when the height \(h\) is \(8\) feet. Substituting \(h = 8\) into the equation gives:

$$8 = 4 + 17t - 16t^2$$

Rearrange into standard form

To solve this quadratic equation, we rearrange it into the standard form \(at^2 + bt + c = 0\).
Subtracting \(8\) from both sides:

$$0 = -16t^2 + 17t - 4$$

Multiplying the entire equation by \(-1\) to make the leading coefficient positive:

$$16t^2 - 17t + 4 = 0$$

Apply the quadratic formula

For a quadratic equation \(at^2 + bt + c = 0\), the solutions are given by:

$$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Here, the coefficients are:

$$a = 16, \quad b = -17, \quad c = 4$$

Substituting these values into the formula:

$$t = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(16)(4)}}{2(16)}$$

$$t = \frac{17 \pm \sqrt{289 - 256}}{32}$$

$$t = \frac{17 \pm \sqrt{33}}{32}$$

Calculate the numerical values

Now we calculate the two possible values for \(t\):
First value:

$$t_1 = \frac{17 + \sqrt{33}}{32} \approx \frac{17 + 5.74456}{32} \approx \frac{22.74456}{32} \approx 0.7108$$

Second value:

$$t_2 = \frac{17 - \sqrt{33}}{32} \approx \frac{17 - 5.74456}{32} \approx \frac{11.25544}{32} \approx 0.3517$$

Round to the nearest hundredth

Rounding each value to the nearest hundredth:

$$t_1 \approx 0.71\text{ seconds}$$

$$t_2 \approx 0.35\text{ seconds}$$

Both values are positive and physically reasonable.

Answer:

\(t =\) <blank>\(0.35\)</blank> or <blank>\(0.71\)</blank> seconds