The height of the ball as it falls can be modeled using the equation of motion under the influence of gravity. The general equation for the height \( h(t) \) of an object in free fall (with no initial velocity and starting from a height \( h_0 \)) is:
\[ h(t) = h_0 - \frac{1}{2} g t^2 \]
where:
- \( h_0 \) is the initial height (in this case, 30 feet),
- \( g \) is the acceleration due to gravity (approximately \( 32 \text{ ft/s}^2 \)),
- \( t \) is the time in seconds.
Given that we are dropping the ball from a height of 30 feet and there is no initial velocity, we can substitute the values into the equation:
- Set \( h_0 = 30 \) feet.
- Use \( g = 32 \text{ ft/s}^2 \).
Thus, the height function becomes:
\[ h(t) = 30 - \frac{1}{2} (32) t^2 \]
This simplifies to:
\[ h(t) = 30 - 16t^2 \]
So the model for the height of the ball from the ground \( h(t) \), in feet, at time \( t \) seconds after it is dropped is:
\[ h(t) = 30 - 16t^2 \]