The ball takes 1.5 seconds to reach its maximum height
The ball's maximum height is 44 feet
Solution:
Given that,
The height in feet after t seconds is given by the function:
[tex]h(t) = -16t^2+48t+8[/tex]
Let us first find the derivative of h(t)
Differentiate the given function with respect to t
[tex]\frac{dh(t)}{dt} = -16 \times 2 \times t+48+0\\\\\frac{dh(t)}{dt} = -32t + 48[/tex]
At the maximum height the derivative of the function needs to be 0
[tex]0=-32t+48\\\\-32t = -48\\\\t = \frac{-48}{-32}\\\\t=1.5[/tex]
Thus the ball takes 1.5 seconds to reach its maximum height
Find the ball's maximum height:
Substitute t = 1.5 in given function
[tex]h=-16(1.5)^2+48(1.5)+8\\\\h = -16 \times 2.25 + 72+8\\\\h = -36 + 80 = 44[/tex]
Thus ball's maximum height is 44 feet
