Answer:
![u = 260.22m/s](https://img.qammunity.org/2022/formulas/engineering/college/rapa72nv4fmdqihbxbdmeqa21s62on7jfj.png)
![S_(max) = 1141.07ft](https://img.qammunity.org/2022/formulas/engineering/college/4qkfrb71l88fgy8puj31ytj6u5bl225dno.png)
Step-by-step explanation:
Given
--- Initial altitude
-- Altitude after 16.5 seconds
--- Acceleration (It is negative because it is an upward movement i.e. against gravity)
Solving (a): Final Speed of the rocket
To do this, we make use of:
![S = ut + (1)/(2)at^2](https://img.qammunity.org/2022/formulas/engineering/college/s7reh0wmda3fhoy29le8poktfwpqbbk5wu.png)
The final altitude after 16.5 seconds is represented as:
![S_(16.5) = S_0 + ut + (1)/(2)at^2](https://img.qammunity.org/2022/formulas/engineering/college/x6zrxncbuwo624id6sszvkij5hm7saqzdm.png)
Substitute the following values:
and
![t = 16.5](https://img.qammunity.org/2022/formulas/engineering/college/sn4afwu647xir0a8xheqctz3i81jzoax1i.png)
So, we have:
![0 = 89.6 + u * 16.5 - (1)/(2) * 32.2 * 16.5^2](https://img.qammunity.org/2022/formulas/engineering/college/nxauow0fzq1viv7gzxao2p58zgefu22ywj.png)
![0 = 89.6 + u * 16.5 - (1)/(2) * 8766.45](https://img.qammunity.org/2022/formulas/engineering/college/ugipgdsoqakp3dw28qrnixoiuwtzlxdegh.png)
![0 = 89.6 + 16.5u- 4383.225](https://img.qammunity.org/2022/formulas/engineering/college/7dl6xg4yh76f1sowr39hn1gfocldr5aihv.png)
Collect Like Terms
![16.5u = -89.6 +4383.225](https://img.qammunity.org/2022/formulas/engineering/college/dgg1sjmj9tab5wd9jk34wl7e6asulvwb5e.png)
![16.5u = 4293.625](https://img.qammunity.org/2022/formulas/engineering/college/37w1ok0sp5z7uiecnta9f2knayhntwwo4p.png)
Make u the subject
![u = (4293.625)/(16.5)](https://img.qammunity.org/2022/formulas/engineering/college/2mnqekf3lz7jhhq45gwsfh6c3rg2iid89u.png)
![u = 260.21969697](https://img.qammunity.org/2022/formulas/engineering/college/fiiid0p51hzd5zumh6x6xpqsdb9acvkl3r.png)
![u = 260.22m/s](https://img.qammunity.org/2022/formulas/engineering/college/rapa72nv4fmdqihbxbdmeqa21s62on7jfj.png)
Solving (b): The maximum height attained
First, we calculate the time taken to attain the maximum height.
Using:
![v=u + at](https://img.qammunity.org/2022/formulas/engineering/college/jz4j1ug9hy5zrd1u5rid62kn2832wmwjg3.png)
At the maximum height:
--- The final velocity
![u = 260.22m/s](https://img.qammunity.org/2022/formulas/engineering/college/rapa72nv4fmdqihbxbdmeqa21s62on7jfj.png)
So, we have:
![0 = 260.22 - 32.2t](https://img.qammunity.org/2022/formulas/engineering/college/c6x3kt7g6h8ulfzf7ris6naaxrclbskq9a.png)
Collect Like Terms
![32.2t = 260.22](https://img.qammunity.org/2022/formulas/engineering/college/xvz7ikv546ozulehayzsvdg66y6ny80xe2.png)
Make t the subject
![t = (260.22)/( 32.2)](https://img.qammunity.org/2022/formulas/engineering/college/wsi2dzorwo9iwe9f1yq21mfnenc2j0znm2.png)
![t = 8.08s](https://img.qammunity.org/2022/formulas/engineering/college/cz0i4v68fw73af3znh8uascjcobuej5ku4.png)
The maximum height is then calculated as:
![S_(max) = S_0 + ut + (1)/(2)at^2](https://img.qammunity.org/2022/formulas/engineering/college/q3yxluvhpkevvq2cbbcqkqtoxmbab4kjlv.png)
This gives:
![S_(max) = 89.6 + 260.22 * 8.08 - (1)/(2) * 32.2 * 8.08^2](https://img.qammunity.org/2022/formulas/engineering/college/za4t6jm329sgqxa0uht164s4h3i95w6vrt.png)
![S_(max) = 89.6 + 260.22 * 8.08 - (1)/(2) * 2102.22](https://img.qammunity.org/2022/formulas/engineering/college/om4smo1qgm9iwlasjqdp9gzgmyfqe8dpcz.png)
![S_(max) = 89.6 + 260.22 * 8.08 - 1051.11](https://img.qammunity.org/2022/formulas/engineering/college/3abp2j0nzin816efqn70poxywbwjveelq6.png)
![S_(max) = 1141.0676](https://img.qammunity.org/2022/formulas/engineering/college/3j3lu4kfzq6vrbbqfch0w2gkcn4jrq5etw.png)
![S_(max) = 1141.07ft](https://img.qammunity.org/2022/formulas/engineering/college/4qkfrb71l88fgy8puj31ytj6u5bl225dno.png)
Hence, the maximum height is 1141.07ft