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

Answer 1

The height \( h(t) \) of a ball thrown directly upwards from the ground can be modeled by the equation:

\[ h(t) = -16t^2 + v_0 t + h_0 \]

where:

• \( v_0 \) is the initial velocity,
• \( h_0 \) is the initial height (in this case, 0 since it is thrown from the ground),
• \( t \) is the time in seconds.

Given that the initial velocity \( v_0 \) is 4.8 ft/s and the initial height \( h_0 \) is 0, the equation simplifies to:

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

Therefore, the appropriate model for the height of the ball from the ground \( t \) seconds after it was thrown upward is:

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