An object is projected vertically upward with an initial velocity of v0 ft/sec, and its distance s(t) in feet above the ground after t seconds is given by the formula s(t) = −16t^2 + v0t.

(a) If the object hits the ground after 8 seconds, find its initial velocity v0.
v0 = ft/sec

(b) Find its maximum distance above the ground.
s(t) = ft

v0 t - 16t^2 = 0 when t=8, so

8 v0 - 1024 = 0
v0 = 128

Now just find the vertex of the parabola, at t = -b/2a

which in this case is at t = v0/32 = 4

that sounds correct, since it is at half the time to hit. 4 sec up, 4 sec down. So, just find s(4)

(a) To find the initial velocity, we can use the given information that the object hits the ground after 8 seconds. We know that when the object hits the ground, its distance above the ground is 0.

Given:
s(t) = -16t^2 + v0t
t = 8 seconds
s(t) = 0

Substituting the values into the equation, we have:
0 = -16(8)^2 + v0(8)

Simplifying the equation:
0 = -16(64) + 8v0
0 = -1024 + 8v0

To solve for v0, we isolate the variable:
8v0 = 1024
v0 = 1024 / 8
v0 = 128 ft/sec

Therefore, the initial velocity of the object is 128 ft/sec.

(b) To find the maximum distance above the ground, we need to determine the vertex of the parabolic function s(t) = -16t^2 + v0t. The vertex can be found using the formula t = -b / 2a, where a = -16 and b = v0.

Given:
a = -16
b = v0 = 128 ft/sec

t = -b / 2a
t = -128 / (2 * (-16))
t = -128 / (-32)
t = 4 seconds

Now, substitute the value of t = 4 seconds into the equation s(t):
s(t) = -16t^2 + v0t
s(4) = -16(4)^2 + 128(4)
s(4) = -16(16) + 512
s(4) = -256 + 512
s(4) = 256 ft

Therefore, the maximum distance above the ground is 256 ft.

To find the initial velocity (v0), we need to use the information that the object hits the ground after 8 seconds.

Given the formula for distance (s(t)) as s(t) = -16t^2 + v0t, we can substitute the given time value of 8 seconds to get:

s(8) = -16(8)^2 + v0(8)

Since the object hits the ground, the distance from the ground is 0. Therefore, we have:

0 = -16(8)^2 + v0(8)

Simplifying the equation:

0 = -16(64) + 8v0
0 = -1024 + 8v0

Adding 1024 to both sides of the equation:

8v0 = 1024

Dividing both sides of the equation by 8:

v0 = 128 ft/sec

Therefore, the initial velocity (v0) of the object is 128 ft/sec.

Now let's move on to part (b) which is finding the maximum distance above the ground.

To find the maximum distance, we need to analyze the equation for s(t) = -16t^2 + v0t. This equation represents a quadratic function with a negative coefficient for the leading term (t^2), which means it is a downward-opening parabola.

For a downward-opening parabola, the maximum value occurs at the vertex of the parabola. The x-coordinate of the vertex can be found using the formula:

x = -b / (2a)

In our case, 'a' is -16 (the coefficient of t^2) and 'b' is v0 (the coefficient of t).

So, substituting the values into the formula:

x = -v0 / (2 * -16)
x = v0 / 32

To find the maximum distance, we substitute this x-coordinate back into the equation for s(t):

s(x) = -16(x)^2 + v0(x)
s(x) = -16(v0/32)^2 + v0(v0/32)
s(x) = -16(v0^2)/32^2 + v0(v0)/32
s(x) = -v0^2/32 + v0^2/32
s(x) = v0^2/32 - v0^2/32
s(x) = 0

So, the maximum distance above the ground is 0 ft. This indicates that the object reaches its maximum height and then falls back to the ground.

Note: The above result may seem counterintuitive, but remember that the equation only represents the vertical distance. The object still moves horizontally due to its initial velocity, but its highest vertical point is at ground level (0 ft).