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 mc001-1.jpg 0.87 s t mc001-2.jpg 4 s t = 8.5 s t = 16 s

Answer 1

$\bf h=-16t^2+255\implies h=-16t^2+0t+255 \\\\\\ ~~~~~~~~~~~~\textit{quadratic formula}\\\\ \begin{array}{lcccl} h=& -16 t^2& +0 t& +255\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \qquad \qquad h= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ h=\cfrac{-0\pm\sqrt{0^2-4(-16)(255)}}{2(-16)}\implies h=\cfrac{\pm\sqrt{16320}}{-32} \\\\\\ h=\cfrac{\pm\sqrt{8^2\cdot 255}}{-32}\implies h=\cfrac{\pm 8\sqrt{255}}{-32}\implies h=\cfrac{\mp \sqrt{255}}{4} \\\\\\ h\approx \mp 3.992179856$

Answer 2

The time it takes the rock to reach the canyon is 4.0 s.

The equation of the object's motion

$h= -16t^2 + 255$

The time it takes the rock to reach the canyon is calculated as follows;

$0 = -16t^2 + 255\\\\16t^2 = 255\\\\t^2 = \frac{255}{16} \\\\t^2 = 15.94\\\\t = \sqrt{15.94} \\\\t = 4.0 \ s$

Thus, the time it takes the rock to reach the canyon is 4.0 s.

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