Question

for this item, enter the answer in the space provided. an engineer creates a robot that is designed to shoot a basketball. the height, h, of the basketball with respect to time, t, is modeled by the function h(t)=−16t² + 20t + 6. a standard basketball hoop is 10 feet high, so the engineer models the height of the hoop with the function g(t)=10. at what time does the basketball reach the hoop as it is falling? second(s)

Explanation:

Step1: Set the two functions equal

$-16t^{2}+20t + 6=10$

Step2: Rearrange to standard quadratic form

$-16t^{2}+20t - 4 = 0$. Divide by -4 to simplify: $4t^{2}-5t + 1=0$

Step3: Use quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 4$, $b=-5$, $c = 1$. So $t=\frac{5\pm\sqrt{(-5)^{2}-4\times4\times1}}{2\times4}=\frac{5\pm\sqrt{25 - 16}}{8}=\frac{5\pm\sqrt{9}}{8}=\frac{5\pm3}{8}$

Step4: Find the two solutions

$t_1=\frac{5 + 3}{8}=1$ and $t_2=\frac{5-3}{8}=\frac{1}{4}$

Step5: Determine the falling - time solution

The ball reaches the hoop on the way up at $t=\frac{1}{4}$ and on the way down at $t = 1$.

Answer:

$1$