Solution:
Given that the height is expressed as
![\begin{gathered} H(t)=48\sin((\pi)/(20)t+(3\pi)/(2))+54\text{ ---- equation 1} \\ where \\ t\text{ is the time i seconds} \\ H(t)\text{ is tghe height in feet} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8sq07022xswtoylsm3wmu66vy2tig9gonr.png)
To find the maximum height,
step 1: Take the first derivative.
Thus,
![H(t)^(\prime)=(12\pi)/(5)\sin\left((\pi t)/(20)\right)---\text{ equation 2}](https://img.qammunity.org/2023/formulas/mathematics/college/2ppiggqeuu643x06w8uqcsue9cgg8rm0uy.png)
step 2: Find the critical point.
At the critical point, H(t)' equals zero.
Thus,
![\begin{gathered} (12\pi)/(5)\sin\left((\pi t)/(20)\right)=0 \\ \Rightarrow\sin\left((\pi t)/(20)\right)=0 \\ take\text{ the sine inverse of both sides} \\ \sin^(-1)(\sin\left((\pi t)/(20)\right))=\sin^(-1)(0) \\ (\pi)/(20)t=\pi \\ thus, \\ t=20 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lwpxkdl13l3br45g7xtt2rrw866gzrmryu.png)
step 3: Take the second derivative.
Thus, we have
![\begin{gathered} H(t)^(\prime)^(\prime)=(3\pi^2\cos\left((\pi t)/(20)\right))/(25) \\ when\text{ t=20,} \\ H(t)^(\prime)^(\prime)=-(3\pi^2)/(25) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5vy8ot2i9a7pvxxuimtpf54tb1cvzx8mtj.png)
Since H(t)'' is negative, we have a maximum point.
To evaluate the maximum height, we substitute the value of 20 for t into the H(t) function.
Thus, we have
![\begin{gathered} H(t)=48\sin((\pi)/(20)t+(3\pi)/(2))+54 \\ t=20 \\ H(20)=48\sin((\pi)/(20)(20)+(3\pi)/(2))+54 \\ =102\text{ feet} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jz40ncjg4xj9hi6lpfslah5ywfkebll16j.png)
Hence, the correct option is