Given Dominic’s initial vertical speed is \( 0 \, \text{m s}^{-1} \) at \( t = 0 \), substitute into \( S(t) = K – 60(1.2^{-t}) \):

\[ S(0) = K – 60(1.2^0) = 0 \]

\[ K – 60 \cdot 1 = 0 \implies K – 60 = 0 \implies K = 60 \]

Answer: \( K = 60 \)

The horizontal asymptote represents the limiting vertical speed Dominic approaches as time increases. To find it, evaluate \( S(t) \) as \( t \to \infty \):

\[ S(t) = 60 – 60(1.2^{-t}) \]

Since \( 1.2^{-t} = \frac{1}{1.2^t} \), as \( t \to \infty \), \( 1.2^t \to \infty \), so \( 1.2^{-t} \to 0 \). Thus:

\[ S(t) \to 60 – 60 \cdot 0 = 60 \]

In the context of the model, this represents the terminal velocity (downward) that Dominic approaches during freefall.

Answer: The horizontal asymptote represents the terminal vertical speed Dominic approaches as time increases, which is \( 60 \, \text{m s}^{-1} \) downward.

Using \( K = 60 \), the speed function is \( S(t) = 60 – 60(1.2^{-t}) \). Find the speed at \( t = 10 \):

\[ S(10) = 60 – 60(1.2^{-10}) \]

Calculate \( 1.2^{-10} \):

\[ 1.2^{-10} = \frac{1}{1.2^{10}} \approx \frac{1}{6.1917364224} \approx 0.161505582 \]

\[ 60 \times 0.161505582 \approx 9.69033492 \]

\[ S(10) = 60 – 9.69033492 \approx 50.30966508 \, \text{m s}^{-1} \]

Convert to \( \text{km h}^{-1} \): \( 1 \, \text{m s}^{-1} = 3.6 \, \text{km h}^{-1} \):

\[ 50.30966508 \times 3.6 \approx 181.114794288 \, \text{km h}^{-1} \]

Rounded to three significant figures:

\[ 181 \, \text{km h}^{-1} \]

Since the speed is downward, the answer is negative in the context of vertical motion (where downward is negative):

Answer: \( -181 \, \text{km h}^{-1} \)