A ball 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

Answer: the maximum heigth was 36 m

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olemakpadu olemakpadu · Answer 2 · 2016-03-16T09:11:13+01:00

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.