The equation $y = -16t^2 + 80t$ describes the height (in feet) of a projectile launched from the ground at 80 feet per second. At what $t$ will the projectile reach 36 feet in height for the first time? Express your answer as a decimal rounded to the nearest tenth.
Setting $y$ to 36, we find the following: \begin{align*}
36& = -16t^2 + 80t\\
0 & = -16t^2 + 80t - 36\\
& = 4t^2 - 20t + 9\\
& = (2t - 1)(2t - 9)
\end{align*}Our possible values for $t$ are $\frac{1}{2} = 0.5$ or $\frac{9}{2} = 4.5.$ Of these, we choose the smaller $t$, or $\boxed{0.5}.$
0.5
Level 4
Algebra
Problem: The equation $y = -16t^2 + 80t$ describes the height (in feet) of a projectile launched from the ground at 80 feet per second. At what $t$ will the projectile reach 36 feet in height for the first time? Express your answer as a decimal rounded to the nearest tenth.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the time \( t \) when the projectile reaches a height of 36 feet, we need to set the height equation equal to 36. This gives us the equation:

\[
-16t^2 + 80t = 36
\]

Rearranging this equation leads to:

\[
-16t^2 + 80t - 36 = 0
\]

Next, we can simplify the equation by dividing all terms by 4:

\[
-4t^2 + 20t - 9 = 0
\]

To solve this quadratic equation, we can apply the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = -4 \), \( b = 20 \), and \( c = -9 \). We first calculate the discriminant:

\[
b^2 - 4ac = 20^2 - 4(-4)(-9) = 400 - 144 = 256
\]

Since the discriminant is positive, there will be two real solutions. Now we compute the square root of the discriminant:

\[
\sqrt{256} = 16
\]

Substituting back into the quadratic formula gives:

\[
t = \frac{-20 \pm 16}{2(-4)}
\]

Calculating the two potential solutions:

1. For the positive root:

\[
t = \frac{-20 + 16}{-8} = \frac{-4}{-8} = 0.5
\]

2. For the negative root:

\[
t = \frac{-20 - 16}{-8} = \frac{-36}{-8} = 4.5
\]

The two solutions are \( t = 0.5 \) and \( t = 4.5 \). Since the problem asks for the first time the projectile reaches 36 feet, we select the smaller value:

The final answer is \( 0.5 \).<|END_OF_LATENT|> The final answer is $0.5$.