Answer:
Step-by-step explanation:
Given
h varies as
[tex]h=-16t^2+64t+190[/tex]
Above h represents the height of object at any instant t
For example at t=0
h=190 ft
At t=1 s
h=-16+64+190=238 ft
For maximum height differentiate h w.r.t time
[tex]\frac{\mathrm{d} h}{\mathrm{d} t}=-16\times 2\times t+64[/tex]
Put [tex]\frac{\mathrm{d} h}{\mathrm{d} t}=0[/tex] to get maximum height
[tex]-32t+64=0[/tex]
t=2 i.e. at t= 2 s h is maximum
h at t=2 s
[tex]h=-16\times 2^2+64\times 2+190=254 m[/tex]
