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)=−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)=−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

Answer 1

To model the height of the ball thrown upward, we can use the following physics concepts:

1. The height \( h(t) \) in feet after \( t \) seconds can be described by the equation: \[ h(t) = -16t^2 + v_0 t + h_0 \] where:

   • The term \(-16t^2\) represents the effect of gravity (in feet per second squared),
   • \( v_0 \) is the initial velocity,
   • \( h_0 \) is the initial height, which in this case is 0 because it is thrown from the ground.

2. Given that the initial velocity \( v_0 \) is 4.8 feet/second, and since the ball is thrown from the ground, the initial height \( h_0 \) is 0.

Thus, the equation becomes: \[ h(t) = -16t^2 + 4.8t \]

Therefore, the correct response is: \[ h(t) = -16t^2 + 4.8t \]