Question

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)
○ h(t)=-\frac{1}{2}cdot4.8t^{2}+4.8t
○ h(t)=-\frac{1}{2}cdot32t^{2}+4.8t + 4.8
○ h(t)=-16t^{2}+4.8t
○ h(t)=-\frac{1}{2}cdot9.8t^{2}+4.8t

Explanation:

Step1: Recall height - time formula

The general formula for the height $h(t)$ of an object in vertical - motion under the influence of gravity is $h(t)=-\frac{1}{2}gt^{2}+v_{0}t + h_{0}$, where $g$ is the acceleration due to gravity, $v_{0}$ is the initial velocity, and $h_{0}$ is the initial height.

Step2: Identify given values

The ball is thrown from the ground, so $h_{0} = 0$. The initial velocity $v_{0}=4.8$ ft/sec. In English units, the acceleration due to gravity $g = 32$ ft/$s^{2}$.

Step3: Substitute values into formula

Substituting $h_{0}=0$, $v_{0}=4.8$ ft/sec, and $g = 32$ ft/$s^{2}$ into $h(t)=-\frac{1}{2}gt^{2}+v_{0}t + h_{0}$, we get $h(t)=-\frac{1}{2}\times32t^{2}+4.8t+0=-16t^{2}+4.8t$.

Answer:

$h(t)=-16t^{2}+4.8t$