Question

A ball is thrown directly upward from the ground with an initial velocity of 4.8 ft./sec. Represent the
height of the ball from the ground t seconds after it was thrown upward using the model h(t). (1 point)
h(t) = − 16t
2 + 4.8t
h(t) = −
1
2
⋅ 32t
2 + 4.8t + 4.8
h(t) = −
1
2
⋅ 4.8t
2 + 4.8t
h(t) = −
1
2
⋅ 9.8t
2 + 4.8t

Answer 1

To model the height of a ball thrown directly upward, we can utilize the general formula for the height \( h(t) \) of an object in free fall:

\[
h(t) = -\frac{1}{2}gt^2 + v_0t + h_0
\]

Where:
- \( g \) is the acceleration due to gravity (in ft/s\(^2\)),
- \( v_0 \) is the initial velocity (in ft/s),
- \( h_0 \) is the initial height (in feet), which is 0 if thrown from the ground.

In this case:
- The initial velocity \( v_0 = 4.8 \) ft/s,
- The initial height \( h_0 = 0 \) ft (since it is thrown from the ground),
- The acceleration due to gravity \( g = 32 \) ft/s\(^2\) (standard value for gravity; it's important to note that when using the formula, it is common to use \( -\frac{1}{2} \cdot g \), hence \( -\frac{1}{2} \cdot 32 = -16 \)).

Thus, we can substitute these values into the model:

\[
h(t) = -16t^2 + 4.8t + 0
\]

So, the height of the ball from the ground \( t \) seconds after it was thrown upward is given by:

\[
h(t) = -16t^2 + 4.8t.
\]

Therefore, the correct representation among the options provided is:

**h(t) = −16t² + 4.8t**.