To represent the height of a ball thrown upward from the ground with an initial velocity, we can use the 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 (approximately 32 ft/s² in feet),
- \( v_0 \) is the initial velocity (4.8 ft/s in this case),
- \( h_0 \) is the initial height (0 ft since the ball is thrown from the ground).
In this case, \( g = 32 \) ft/s², \( v_0 = 4.8 \) ft/s, and \( h_0 = 0 \). Plug these values into the formula:
\[ h(t) = -\frac{1}{2}(32)t^2 + 4.8t + 0 \] \[ h(t) = -16t^2 + 4.8t \]
So the correct representation of the height of the ball from the ground \( t \) seconds after it was thrown upward is:
\[ h(t) = -16t^2 + 4.8t \]
This matches the first response option you provided.