a soccer ball is kicked into the air from the ground. if the ball reaches a maximum height of 25 ft and…

Answer:

A. $h(t)=-16t^{2}+40t$

Explanation:

Step1: Recall the general height - time formula

The general formula for the height of an object in vertical - motion under the influence of gravity is $h(t)=at^{2}+vt + h_{0}$, where $a$ is the acceleration due to gravity, $v$ is the initial velocity, and $h_{0}$ is the initial height. Given $a=-16$ ft/s² and $h_{0} = 0$ (kicked from the ground, so $h_{0}=0$), the formula becomes $h(t)=-16t^{2}+vt$.

Step2: Use the time - of - flight information

The time of flight $T = 2.5$ s. The time it takes to reach the maximum height is half of the total time of flight. So the time to reach the maximum height $t_{max}=\frac{T}{2}=\frac{2.5}{2}=1.25$ s.

Step3: Use the maximum - height information

At the maximum height, the derivative of the height function $h(t)$ with respect to time $t$ is 0. The derivative $h^\prime(t)=-32t + v$. At $t = 1.25$ s, $h^\prime(1.25)=0$. So, $-32\times1.25+v = 0$. Solving for $v$: [ \begin{align*} -40 + v&=0\ v&=40 \end{align*} ]

Step4: Write the height function

Substitute $v = 40$ and $h_{0}=0$ into $h(t)=-16t^{2}+vt+h_{0}$, we get $h(t)=-16t^{2}+40t$.