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)

Question

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

Answer 1

The height of an object thrown vertically upward can be modeled by the equation:

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

where:

• \( h(t) \) is the height of the object at time \( t \),
• \( v_0 \) is the initial velocity,
• \( h_0 \) is the initial height (which is 0 in this case since the ball is thrown from the ground),
• The term \(-16t^2\) accounts for the acceleration due to gravity (in feet per second squared).

In this case:

• The initial velocity \( v_0 \) is 4.8 ft/sec,
• The initial height \( h_0 \) is 0 ft.

Substituting the values we have:

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

So, the correct representation of the height of the ball from the ground \( t \) seconds after it is thrown upward is:

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

This corresponds to the first option in your list.