Question

A ball is dropped from a 30-foot-tall building, meaning it has no initial velocity. Write a model h(t) that represents the height of the ball from the ground, in feet, t seconds after it is dropped from the building.

h(t)=__t^(2)+__t+__

Answer 1

When a ball is dropped from a height with no initial velocity, its height as a function of time can be modeled using the following kinematic equation for an object under the influence of gravity:

\[ h(t) = h_0 - \frac{1}{2} g t^2 \]

Where:
- \( h(t) \) is the height above the ground in feet at time \( t \) seconds,
- \( h_0 \) 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.

Plugging in the initial height and the value for \( g \):

\[ h(t) = 30 - \frac{1}{2} \cdot 32 \cdot t^2 \]
\[ h(t) = 30 - 16t^2 \]

Thus, you can express it in the required format \( h(t) = __t^2 + __t + __ \) as follows:

1. The coefficient of \( t^2 \) is \(-16\),
2. The coefficient of \( t \) is \(0\) (because there is no initial velocity),
3. The constant term is \(30\).

So, the model \( h(t) \) can be written as:

\[ h(t) = -16t^2 + 0t + 30 \]

Thus, to fill in the blanks:

h(t) = \(-16t^2 + 0t + 30\)