Aball was kicked into the air from a balcony 20 feet above the ground, and the ball's height above the ground, in feet, t seconds after the ball wasw kicked was h(t) = 20 −16 t^2 + 32t. what was the maximum height, in feet, of the ball above the ground after it was kicked?

1) we have to calculate the first derivative:
 h´(t)=-32t+32

2)the first derivative equate to "0"
-32t+32=0

3) obtein value of "t"
-32t+32=0
t=-32/32=1

4) we have to calculate the second derivative
h´´(T)=-32<0    ⇒  exist a maximum in t=1

5) now calculate the maximum height.

h(1)=20-16(1)²+32(1)=36 m

the maximum heigth was 36 m

H(t) = 20 - 16t² + 32t

Using the differential function for y = axⁿ,  dy/dx =  naxⁿ⁻¹

At maximum height  dh/dt = 0, and also d²h/dt² < 0

dh/dt = 0 - 2*16t²⁻¹ +  1*32t¹⁻¹

dh/dt = 0 - 32t¹ + 32t⁰ =  -32t + 32

dh/dt =  -32t + 32 = 0

-32t + 32 = 0

-32t = 0- 32

-32t = -32

t = -32/-32 = 1

To actually test that at t = 1, that h(t) is at maximum

dh/dt =  -32t + 32 

d²h/dt² = -1*32t¹⁻¹ + 0

d²h/dt² = -1*32t⁰ = -32,

Since d²h/dt² < 0, h(t) has a maximum point.

The function is maximum at t = 1.

h(t) = 20 - 16t² + 32t,      substituting t = 1

h(1) = 20 - 16(1)² + 32(1) = 20 - 16 + 32 = 36

Hence the maximum height  = 36 feet.