At time $t=0,$ a ball is thrown downward at 24 feet per second from a height of 160 feet above the ground.  The equation $h = -16t^2 - 24t +160$ describes the height (in feet) of the ball. In how many seconds will the ball hit the ground? Express your answer as a decimal.
Setting $h$ to zero, we find the following: \begin{align*}
0& = -16t^2 - 24t + 160\\
& = 2t^2 +3t - 20\\
& = (2t-5)(t+4)\\
\end{align*}The negative value of $t$ is extraneous, so we are left with $t=\boxed{2.5}$
2.5
Level 4
Algebra
Problem: At time $t=0,$ a ball is thrown downward at 24 feet per second from a height of 160 feet above the ground.  The equation $h = -16t^2 - 24t +160$ describes the height (in feet) of the ball. In how many seconds will the ball hit the ground? Express your answer as a decimal.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find when the ball hits the ground, set the height equation $h = -16t^2 - 24t + 160$ equal to zero. This represents the moment when the ball reaches ground level. 

The equation becomes $0 = -16t^2 - 24t + 160$. Rearranging gives $16t^2 + 24t - 160 = 0$. 

Next, apply the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 16$, $b = 24$, and $c = -160$. 

Calculate the discriminant: $b^2 - 4ac = 24^2 - 4(16)(-160) = 576 + 10240 = 10816$. 

Now, find the square root of the discriminant: $\sqrt{10816} = 104$. 

Substituting back into the quadratic formula gives $t = \frac{-24 \pm 104}{32}$. 

Calculate the two potential solutions: 

1. $t = \frac{80}{32} = 2.5$ (valid solution, as time cannot be negative).
2. $t = \frac{-128}{32} = -4$ (not valid, as time cannot be negative).

Thus, the time when the ball hits the ground is $t = 2.5$ seconds.<|END_OF_LATENT|> The final answer is $2.5$.