Question

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.
$h = 2 + 27t - 16t^2$
find all values of t for which the balls height is 12 feet.
round your answer(s) to the nearest hundredth.
(if there is more than one answer, use the \or\ button.)

Explanation:

Step1: Set h = 12 in the equation

We know the height equation is \( h = 2 + 27t - 16t^2 \). When \( h = 12 \), we substitute it into the equation:
\( 12 = 2 + 27t - 16t^2 \)

Step2: Rearrange into standard quadratic form

Subtract 12 from both sides to get a quadratic equation in the form \( ax^2+bx + c = 0 \):
\( - 16t^2+27t + 2 - 12=0 \)
Simplify to:
\( - 16t^2+27t - 10 = 0 \)
Multiply both sides by - 1 to make the coefficient of \( t^2 \) positive:
\( 16t^2-27t + 10 = 0 \)
Here, \( a = 16 \), \( b=-27 \), \( c = 10 \)

Step3: Use quadratic formula \( t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \)

First, calculate the discriminant \( D=b^2-4ac \):
\( D=(-27)^2-4\times16\times10=729 - 640 = 89 \)
Then, find the values of \( t \):
\( t=\frac{27\pm\sqrt{89}}{2\times16}=\frac{27\pm9.43398}{32} \)
For the plus sign:
\( t_1=\frac{27 + 9.43398}{32}=\frac{36.43398}{32}\approx1.14 \)
For the minus sign:
\( t_2=\frac{27-9.43398}{32}=\frac{17.56602}{32}\approx0.55 \)

Answer:

\( t = 0.55 \) or \( t = 1.14 \) seconds