Exercise 28 Page 587

We want to determine how long it takes for riders to drop $60$ feet. Let's first find when the Demon Drop reaches to the top of the tower. To do so, we need to find the vertex of the given quadratic function. We need to identify $a\text{-},$ $b\text{-},$ and $c\text{-}$values of the quadratic function. We see that $a=\text{-}16,$ $b=64,$ and $c=\text{-}60.$ We can now find the $t\text{-}$ and $h\text{-}$coordinates of the vertex. Let's find the $t\text{-}$coordinate of the vertex. To find the second coordinate, we need to substitute $2$ for $t$ in the function. The vertex is $(2,4).$ This means that after $2$ seconds the Demon Drop will reach its maximum height, $4$ feet above the $t\text{-}$axis. From that height, it will drop $60$ feet. Let's draw a diagram to see what is happening. Therefore, we need to find when the equation is equal to $\text{-}56$ since this is where the bottom of the ride is located. To do so, we need to solve the equation below. We will solve it by using the Quadratic Formula. We first rewrite it and then identify the values of $a,$ $b,$ and $c.$ We see that $a=\text{-}16,$ $b=64,$ and $c=\text{-}4.$ Let's substitute these values into the Quadratic Formula. The solutions for this equation are $t=\frac{\text{-}64\pm \sqrt{3840}}{\text{-}32}.$ Let's separate them into the positive and negative cases.

$t=\frac{\text{-}64\pm \sqrt{3840}}{\text{-}32}$
$t_1=\frac{\text{-}64+\sqrt{3840}}{\text{-}32}$	$t_2=\frac{\text{-}64-\sqrt{3840}}{\text{-}32}$
$t_1=\frac{64}{32}-\frac{16\sqrt{15}}{32}$	$t_1=\frac{64}{32}+\frac{16\sqrt{15}}{32}$
$t_1\approx 0.1$	$t_2\approx 3.9$

The solutions to the equation are $t_1=0.1$ and $t_2=3.9.$ These solutions means that the Demon Drop is at the height of $\text{-}56$ feet at time $0.1$ and $3.9.$ Hence, it takes $1.9$ seconds for riders to drop $60$ feet from the maximum height $(2,4).$