Solution:
\( v(t)=\frac{dy}{dt}=24-1.6t \)
Set equal to zero to find time of maximum height
\( 24-1.6t=0 \)
\( 1.6t=24 \)
\( t=15s \)
Now plug \( t=15 \) back into our height function to find the maximum height.
\( y(15)=24(15)-0.8(15)^2=360-0.8(225)=360-180=180m \)