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:
- The coefficient of \( t^2 \) is \(-16\),
- The coefficient of \( t \) is \(0\) (because there is no initial velocity),
- 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) = \(-16t^2 + 0t + 30\)
1 answer