PART A:
The zeros of the function are the values of t where we have h = 0.
Looking at the table, the zeros are t = 0 and t = 10.5
PART B:
Let's write the function using the factored form of the quadratic equation:
![h(t)=c(t-t_1)(t-t_2)_{}](https://img.qammunity.org/2023/formulas/physics/college/himqklv4y2qaot2jidjouxwllmvkfiucl8.png)
Since the zeros are 0 and 10.5, we have t1 = 0 and t2 = 10.5, so:
![h(t)=c\cdot t\cdot(t-10.5)](https://img.qammunity.org/2023/formulas/physics/college/1tg51tzgf3r4t0yubdqteyoap0jbaz2156.png)
Now, using the point (2, 272) from the table, we have:
![\begin{gathered} 272=c\cdot2\cdot(2-10.5) \\ 272=2c\cdot(-8.5) \\ -17c=272 \\ c=-16 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/x52m4dkzqixxwil0oysi1n90yyrwmgaqc7.png)
So our function is:
![h(t)=-16\cdot t\cdot(t-10.5)](https://img.qammunity.org/2023/formulas/physics/college/81zxctswwwsf60lso1q5sz9vrlxm0wo4yh.png)
PART C:
First let's expand our function to the standard form:
![\begin{gathered} h(t)=-16t(t-10.5) \\ h(t)=-16t^2+168t+0 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/naq69sr7x63n1xm0a1go17d2h3z0jtknlh.png)
Comparing with the given model, we have V0 = 168 ft/s (because the coefficient multiplying the variable t is 168 in our function and V0 in the given model).
PART D:
First, let's calculate the average value between the zeros:
![t_v=(0+10.5)/(2)=5.25](https://img.qammunity.org/2023/formulas/physics/college/dfje1ckoox7997kn2g3q4vrqou8le8litu.png)
Using this value of t in the function, we have:
![\begin{gathered} h(t_v)=-16\cdot5.25\cdot(5.25-10.5) \\ h(t_v)=441\text{ ft} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/corkthl2vq7qv068c31a21txltuywe0ie0.png)
So the maximum height is 441 ft.