Question

a ball is thrown from an initial height of 2 feet with an initial upward h = 2 + 18t - 16t^2. find all values of t for which the balls height is 6 feet. round your answer(s) to the nearest hundredth. (if there is more than one answer, use the \or\ button.)

Explanation:

Step1: Set up the equation

Set $h = 6$ in the equation $h=2 + 18t-16t^{2}$, so we get $6=2 + 18t-16t^{2}$.

Step2: Rearrange to standard quadratic form

Rearrange the equation to $16t^{2}-18t + 4=0$. Divide through by 2 to simplify: $8t^{2}-9t + 2=0$.

Step3: Apply quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 8$, $b=-9$, $c = 2$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-9)^{2}-4\times8\times2=81 - 64 = 17$. Then $t=\frac{9\pm\sqrt{17}}{16}$.

Step4: Calculate the values of t

$t_1=\frac{9+\sqrt{17}}{16}\approx\frac{9 + 4.123}{16}=\frac{13.123}{16}\approx0.82$ and $t_2=\frac{9-\sqrt{17}}{16}\approx\frac{9 - 4.123}{16}=\frac{4.877}{16}\approx0.30$.

Answer:

$t\approx0.30$ or $t\approx0.82$