Question

the height h (in feet) of an object t seconds after it is dropped can be modeled by the quadratic equation h = -16t² + h₀, where h₀ is the initial height of the object. suppose a small rock dislodges from a ledge that is 255 ft above a canyon floor. solve the equation h = -16t² + 255 for t, using the quadratic formula to determine the time it takes the rock to reach the canyon floor.\
\
\\( t \approx 0.87 \\) s\
\\( t \approx 4 \\) s\
\\( t = 8.5 \\) s\
\\( t = 16 \\) s\
done

Explanation:

Step1: Determine when the rock reaches the canyon floor

When the rock reaches the canyon floor, the height \( h = 0 \). So we set up the equation:

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

Rearrange it to the standard quadratic form \( ax^{2}+bx + c = 0 \). Here, \( a=- 16 \), \( b = 0 \) (since there is no \( t \) term), and \( c = 255 \). The quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \).

Step2: Substitute values into the quadratic formula

Substitute \( a=-16 \), \( b = 0 \), and \( c = 255 \) into the quadratic formula:

$$ t=\frac{-0\pm\sqrt{0^{2}-4\times(-16)\times255}}{2\times(-16)}=\frac{\pm\sqrt{0 + 16320}}{-32}=\frac{\pm\sqrt{16320}}{-32} $$

We can simplify \( \sqrt{16320}\approx127.75 \) (since \( 127^{2}=16129 \) and \( 128^{2}=16384 \), so it's close to 127.75). Since time cannot be negative in this context (we are looking for the time after the rock is dropped), we take the positive root:

$$ t=\frac{- 127.75}{-32}\approx4 $$

Answer:

B. \( t\approx4 \) s