Question

the height (h) (in feet) of an object (t) seconds after it is dropped can be modeled by the quadratic equation (h = -16t^2 + h_0), where (h_0) is the initial height of the object. suppose a small rock dislodges from a ledge that is (255\text{ ft}) above a canyon floor. solve the equation (h = -16t^2 + 255) for (t), using the quadratic formula to determine the time it takes the rock to reach the canyon floor.

(t approx 0.87\text{ s})
(t approx 4\text{ s})
(t = 8.5\text{ s})
(t = 16\text{ s})

Explanation:

⚡ Using what you learned: quadratic formula and its applications

Step 1: Set up the equation

To find when the rock reaches the canyon floor, set the height \( h = 0 \):

$$ 0 = -16t^2 + 255 $$

Step 2: Apply the quadratic formula

Identify the coefficients for the standard quadratic form \( at^2 + bt + c = 0 \):

• \( a = -16 \)
• \( b = 0 \)
• \( c = 255 \)

Substitute these values into the quadratic formula:

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

$$ t = \frac{0 \pm \sqrt{0^2 - 4(-16)(255)}}{2(-16)} $$

$$ t = \frac{\pm \sqrt{16320}}{-32} $$

Step 3: Calculate the positive time value

Since time \( t \) must be positive:

$$ t = \frac{-\sqrt{16320}}{-32} $$

$$ t \approx \frac{-127.75}{-32} $$

$$ t \approx 3.99 \text{ s} \approx 4 \text{ s} $$

Answer:

\( t \approx 4\text{ s} \)