a ball is thrown from an initial height of 2 feet with an initial upward velocity of 27 ft/s. the balls…

a ball is thrown from an initial height of 2 feet with an initial upward velocity of 27 ft/s. the balls height h (in feet) after t seconds is given by the following.\n\n$h = 2+27t - 16t^{2}$\n\nfind all values of t for which the balls height is 12 feet.\n\nround your answer(s) to the nearest hundredth.\n(if there is more than one answer, use the \or\ button.)

a ball is thrown from an initial height of 2 feet with an initial upward velocity of 27 ft/s. the balls height h (in feet) after t seconds is given by the following.\n\n$h = 2+27t - 16t^{2}$\n\nfind all values of t for which the balls height is 12 feet.\n\nround your answer(s) to the nearest hundredth.\n(if there is more than one answer, use the \or\ button.)

Answer

Explanation:

Step1: Set up the equation

Set $h = 12$ in the equation $h=2 + 27t-16t^{2}$. So we get $12=2 + 27t-16t^{2}$.

Step2: Rearrange to standard quadratic - form

Rearrange the equation to $16t^{2}-27t + 10 = 0$.

Step3: Use the quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 16$, $b=-27$, and $c = 10$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-27)^{2}-4\times16\times10=729 - 640 = 89$.

Step4: Find the values of t

$t=\frac{27\pm\sqrt{89}}{32}$. $t_1=\frac{27+\sqrt{89}}{32}\approx\frac{27 + 9.43}{32}=\frac{36.43}{32}\approx1.14$. $t_2=\frac{27-\sqrt{89}}{32}\approx\frac{27 - 9.43}{32}=\frac{17.57}{32}\approx0.55$.

Answer:

$t\approx0.55$ or $t\approx1.14$