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 ft above the 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 approx 8.5 , \text{s} )
• ( t = 16 , \text{s} )

Explanation:

Step1: Set h = 0 (canyon floor)

We have the equation \( 0 = -16t^2 + 255 \). Rearranging it gives \( 16t^2 = 255 \).

Step2: Solve for \( t^2 \)

Divide both sides by 16: \( t^2=\frac{255}{16} \approx 15.9375 \).

Step3: Take square root

Take the square root of both sides. Since time can't be negative, we consider the positive root: \( t=\sqrt{\frac{255}{16}}=\frac{\sqrt{255}}{4}\approx\frac{15.9687}{4}\approx 4 \) (approximate value, or more accurately, using quadratic formula: the quadratic is \( - 16t^2+255 = 0\), which is \( at^2 + bt + c = 0\) with \( a=-16\), \( b = 0\), \( c = 255\). The quadratic formula is \( t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Plugging in, we get \( t=\frac{0\pm\sqrt{0 - 4\times(-16)\times255}}{2\times(-16)}=\frac{\pm\sqrt{16320}}{-32}\). We take the positive root (since time is positive), \( t=\frac{\sqrt{16320}}{32}\approx\frac{127.75}{32}\approx 4 \) (approximate, and among the options, \( t\approx4\) s is the closest).

Answer:

\( t\approx4 \) s (the option: \( t\approx4 \) s)