You throw a stone from a height of 16 feet with an initial vertical velocity of 32 feet per second. the function h= -16t squared + 32t + 16 represents the h ( in feet) of the stone after t seconds) find the maximum height of the stone the maximum height. me somebody teacher expert anyone

hope this helps :)

Step-by-step explanation:

dh/dt=-32t+32

When velocity, dh/dt, is equal to zero, the object is at its maximum height.

dh/dt=0 only when 32t=32, t=1 second, so the maximum height is h(1)

h(1)=-16+32=16ft

Using algebra

The maximum height will occur at the midpoint between the two zeros of the height function because of a quadratics symmetry...

h(t)=0 when

-16t^2+32t=0

-16t(t-2)=0

so t=0 and 2

The midpoint is t=1

And of course this will give you the same as we found earlier.  The maximum height is h(1) ft

dh/dt=-32t+32When velocity, dh/dt, is equal to zero, the object is at its maximum height.dh/dt=0 only when 32t=32, t=1 second, so the maximum height is h(1)h(1)=-16+32=16ftUsing algebraThe maximum height will occur at the midpoint between the two zeros of the height function because of a quadratics symmetry...h(t)=0 when-16t^2+32t=0-16t(t-2)=0so t=0 and 2The midpoint is t=1

Hope this helped Wilbert!

answer: y = -2.5x - 15

              (-6,0)

step-by-step explanation: