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) Responses h(t)=−16t2+4.8t h left parenthesis t right parenthesis equals negative 16 t squared plus 4.8 t h(t)=−12⋅9.8t2+4.8t h left parenthesis t right parenthesis equals negative Start Fraction 1 over 2 End Fraction times 9.8 t squared plus 4.8 t h(t)=−12⋅32t2+4.8t+4.8 h left parenthesis t right parenthesis equals negative Start Fraction 1 over 2 End Fraction times 32 t squared plus 4.8 t plus 4.8 h(t)=−12⋅4.8t2+4.8t

1 answer

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.