Given equation is
[tex] h(t) = -16t^2 + 64t + 80 [/tex]
We have to find the time interval by which height is maximum.
We start with differentiating .
Differentiating h(t) ,
[tex] h'(t) = -16(2)t^{2-1} +64(1) t^{1-1} + 0
\\ = -32t +64 [/tex]
Differentiating again
[tex] h''(t)= -32(1)t^{1-1} +0 = -32 <0 [/tex]
Since on double differentiating , we are getting a negative number, so the height is maximum.
And to find the time of maximum height, we set h'(t) to 0, that is
[tex] -32t + 64=0
\\
32t = 64
\\
t= 2 [/tex]
In second part, the height is decreasing means slope, h'(t) is less than 0. That is,
[tex] -32t+64<0
\\
-32t<-64
\\
t>2 seconds [/tex]
So the required time interval is
[tex] (2 , \infty) [/tex]
