Question

The last answer is 1.5s and 56ft PLZ HELP PLZ

Answer 1

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:

$h(t) = -16t^2+48t+8$

Let us first find the derivative of h(t)

Differentiate the given function with respect to t

$\frac{dh(t)}{dt} = -16 \times 2 \times t+48+0\\\\\frac{dh(t)}{dt} = -32t + 48$

At the maximum height the derivative of the function needs to be 0

$0=-32t+48\\\\-32t = -48\\\\t = \frac{-48}{-32}\\\\t=1.5$

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

$h=-16(1.5)^2+48(1.5)+8\\\\h = -16 \times 2.25 + 72+8\\\\h = -36 + 80 = 44$

Thus ball's maximum height is 44 feet