a baseball is thrown into the air from a height of 5 feet. the ball reaches a maximum height of 43.5 feet…

Answer:

C. $h(t)= - 16t^{2}+49.64t + 5$

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 $h(t)=-\frac{1}{2}gt^{2}+v_{0}t + h_{0}$, where $g$ is the acceleration due to gravity, $v_{0}$ is the initial velocity, and $h_{0}$ is the initial height. Given $g = 32$ ft/s², so $-\frac{1}{2}g=-16$ ft/s², and $h_{0}=5$ feet. So the equation is of the form $h(t)=-16t^{2}+v_{0}t + 5$.

Step2: Use the time - of - flight information

The time of flight $T$ of a projectile is given by the formula $T=\frac{-v_{0}\pm\sqrt{v_{0}^{2}-4(-16)(h_{0})}}{-32}$ (from the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $h(t)=-16t^{2}+v_{0}t + h_{0}$, where $a=-16$, $b = v_{0}$, $c = h_{0}$). We know $T = 3.2$ s and $h_{0}=5$ ft. Using the fact that for a projectile launched and landing at the same height (in this case, we can use the time - of - flight formula), $T=\frac{v_{0}}{16}$ (when the starting and ending heights are the same, the quadratic formula simplifies for the non - zero root). Since $T = 3.2$ s, then $v_{0}=16\times3.1 = 49.6$ ft/s. So the equation is $h(t)=-16t^{2}+49.64t + 5$.