the height h (in feet) of an object t seconds after it is dropped can be modeled by the quadratic equation h…
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 ≈ 0.87 s t ≈ 4 s t = 8.5 s t = 16 s
Answer
Explanation:
Step1: Substitute (h = 0) into the equation
When the rock reaches the canyon floor, (h = 0). So the equation becomes (0=-16t^{2}+255). Rearranging it to the standard quadratic form (ax^{2}+bx + c = 0), we get (16t^{2}-255 = 0), where (a = 16), (b = 0), and (c=-255).
Step2: Apply the quadratic formula (t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a})
Substitute (a = 16), (b = 0), and (c=-255) into the formula: [ \begin{align*} t&=\frac{-0\pm\sqrt{0^{2}-4\times16\times(-255)}}{2\times16}\ &=\frac{\pm\sqrt{4\times16\times255}}{32}\ &=\frac{\pm\sqrt{16320}}{32}\ &=\frac{\pm127.75}{32} \end{align*} ] Since time (t>0), we take the positive root (t=\frac{127.75}{32}\approx4)
Answer:
(t\approx4\ s)