To solve the problem, we start by using the given formula for the height of an object dropped from a height \( s \):
\[ h = -16t^2 + s \]
In this scenario, \( s = 18 \) feet (the height from which the egg is dropped) and we are looking for the time \( t \) when the egg hits the ground. At that point, the height \( h \) will be \( 0 \) feet.
We set the formula equal to zero to find \( t \):
\[ 0 = -16t^2 + 18 \]
Next, we rearrange the equation to solve for \( t \):
\[ 16t^2 = 18 \]
\[ t^2 = \frac{18}{16} \]
\[ t^2 = \frac{9}{8} \]
Now, we take the square root of both sides to solve for \( t \):
\[ t = \sqrt{\frac{9}{8}} = \frac{3}{\sqrt{8}} = \frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{4} \]
To find a numerical value for \( t \), we can calculate it approximately:
\[ t \approx \frac{3 \times 1.414}{4} = \frac{4.242}{4} \approx 1.0605 \]
Rounding that to the nearest tenth of a second, we find:
\[ t \approx 1.1 \text{ seconds} \]
Therefore, it took approximately 1.1 seconds for the egg to hit the ground.