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)=−16t2+4.8t
h left parenthesis t right parenthesis equals negative 16 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⋅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
Skip to navigation

page 14 of 14

Answer 1

To model the height of a ball thrown upward from the ground, we can use the quadratic equation:

\[ h(t) = -16t^2 + v_0t + h_0 \]

where:

• \( h(t) \) is the height at time \( t \),
• \( -16 \) represents the acceleration due to gravity in feet per second squared (the factor of -16 comes from \(-\frac{1}{2} \cdot 32\), where \(32\) is the acceleration due to gravity),
• \( v_0 \) is the initial velocity (in feet per second), and
• \( h_0 \) is the initial height (with \( h_0 = 0 \) since it was thrown from the ground).

Given an initial velocity of \( 4.8 \) ft/sec, and since the ball is thrown from ground level, the equation simplifies to:

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

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

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

This corresponds to the first option provided:

h(t)=−16t^2+4.8t