Solution:
Given that the height is expressed as
[tex]\begin{gathered} H(t)=48\sin(\frac{\pi}{20}t+\frac{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}[/tex]
To find the maximum height,
step 1: Take the first derivative.
Thus,
[tex]H(t)^{\prime}=\frac{12\pi}{5}\sin\left(\frac{\pi t}{20}\right)---\text{ equation 2}[/tex]
step 2: Find the critical point.
At the critical point, H(t)' equals zero.
Thus,
[tex]\begin{gathered} \frac{12\pi}{5}\sin\left(\frac{\pi t}{20}\right)=0 \\ \Rightarrow\sin\left(\frac{\pi t}{20}\right)=0 \\ take\text{ the sine inverse of both sides} \\ \sin^{-1}(\sin\left(\frac{\pi t}{20}\right))=\sin^{-1}(0) \\ \frac{\pi}{20}t=\pi \\ thus, \\ t=20 \end{gathered}[/tex]
step 3: Take the second derivative.
Thus, we have
[tex]\begin{gathered} H(t)^{\prime}^{\prime}=\frac{3\pi^2\cos\left(\frac{\pi t}{20}\right)}{25} \\ when\text{ t=20,} \\ H(t)^{\prime}^{\prime}=-\frac{3\pi^2}{25} \end{gathered}[/tex]
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
[tex]\begin{gathered} H(t)=48\sin(\frac{\pi}{20}t+\frac{3\pi}{2})+54 \\ t=20 \\ H(20)=48\sin(\frac{\pi}{20}(20)+\frac{3\pi}{2})+54 \\ =102\text{ feet} \end{gathered}[/tex]
Hence, the correct option is
