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)=−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
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

Answer 1

To model the height of a ball thrown directly upward from the ground, we can use the following physics equation for vertical motion:

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

where:

• \( h(t) \) is the height of the ball at time \( t \),
• \( g \) is the acceleration due to gravity (32 ft/s² for feet),
• \( v_0 \) is the initial velocity.

Given that the initial velocity \( v_0 = 4.8 \) ft/s, we can plug this into our equation. The correct value of \( g \) in feet per second squared is 32 ft/s², so we have:

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

Simplifying this gives:

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

Thus, the correct response among the options is:

h(t) = -16t² + 4.8t.