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)=-16t² + 4.8t h(t)=-1/2·32t² + 4.8t + 4.8 h(t)=-1/2·4.8t² + 4.8t h(t)=-1/2·9.8t² + 4.8t

Explanation:

Step1: Recall the height - time formula

The general formula for the height $h(t)$ of an object in vertical - motion under the influence of gravity is $h(t)=h_0 + v_0t-\frac{1}{2}gt^2$, where $h_0$ is the initial height, $v_0$ is the initial velocity, and $g$ is the acceleration due to gravity.

Step2: Identify the initial conditions

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 the values into the formula

Substituting $h_0 = 0$, $v_0 = 4.8$ ft/sec, and $g = 32$ ft/$s^2$ into $h(t)=h_0 + v_0t-\frac{1}{2}gt^2$, we get $h(t)=0 + 4.8t-\frac{1}{2}\times32t^2=-16t^2 + 4.8t$.

Answer:

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