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)

Question

.(1 point)
Responses

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⋅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⋅4.8t2+4.8t
h left parenthesis t right parenthesis equals negative Start Fraction 1 over 2 End Fraction times 4.8 t squared plus 4.8 t

h(t)=−16t2+4.8t

Answer 1

To model the height of a ball thrown directly upward from the ground, we use the equation of motion given by:

\[ h(t) = -\frac{1}{2} g t^2 + v_0 t + h_0 \]

where:

• \(g\) is the acceleration due to gravity (32 ft/sec² when using feet),
• \(v_0\) is the initial velocity,
• \(h_0\) is the initial height (which is 0 if thrown from the ground).

In this scenario:

• The initial velocity \(v_0\) is \(4.8\) ft/sec,
• The initial height \(h_0\) is \(0\) ft,
• The standard value for gravity \(g\) is \(32\) ft/sec².

Substituting these values into the equation, we get:

\[ h(t) = -\frac{1}{2} \times 32 t^2 + 4.8 t + 0 \] \[ h(t) = -16t^2 + 4.8t \]

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

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

Therefore, the correct response is:

h(t) = -16t² + 4.8t