Question

for the following question, use the function (h = -16t^2 + v_0t + h_0), where (h) is the height of the object after time (t), (v_0) is the initial velocity, and (h_0) is the initial height.

an object is thrown upward from a height of (240\text{ ft}) with an initial velocity of (32\text{ ft/s}). how long will it take for the object to reach the ground?

Explanation:

Identify the given values and formula

We are given the height function:

$$h = -16t^2 + v_0t + h_0$$

From the problem statement, we identify the following values:

• Initial height, \(h_0 = 240\text{ ft}\)
• Initial velocity, \(v_0 = 32\text{ ft/s}\)

We want to find how long it takes for the object to reach the ground, which means finding the time \(t\) when the height \(h = 0\).

Set up the quadratic equation

Using the Quadratic Functions concept, we substitute the given values into our height equation:

$$0 = -16t^2 + 32t + 240$$

Simplify the equation

To make solving easier, we can divide the entire equation by its greatest common divisor, \(-16\):

$$\frac{-16t^2}{-16} + \frac{32t}{-16} + \frac{240}{-16} = \frac{0}{-16}$$

$$t^2 - 2t - 15 = 0$$

Factor the quadratic equation

Using the Solving Quadratic Equations concept, we factor the simplified quadratic expression:
We look for two numbers that multiply to \(-15\) and add to \(-2\). These numbers are \(-5\) and \(3\).

$$(t - 5)(t + 3) = 0$$

Solve for time

Set each factor to zero:

$$t - 5 = 0 \implies t = 5$$

$$t + 3 = 0 \implies t = -3$$

Since time \(t\) must be a non-negative value in this context, we discard the negative solution.
Thus, \(t = 5\text{ seconds}\).

Answer:

How long will it take for the object to reach the ground? <blank>5</blank> s