An object is projected vertically upward from the top of a building with an initial velocity of 144 ft/sec. Its distance s(t) in feet above the ground after t seconds is given by the function

s(t) = −16t2 + 144t + 120.
A) Find its maximum distance above the ground. (in feet)
B) Find the height of the building. (in feet)

v(t) = -32t + 144

at max height v(t) = 0
32t = 144
t = 4.5
s(4.5) = -16(4.5)^2 + 144(4.5) + 120 = 444 ft

b) height of building --- t = 0
height = 120

A) To find the maximum distance above the ground, we need to determine the vertex of the parabolic function. The vertex is the highest point on the graph, which corresponds to the maximum distance.

The formula for finding the x-coordinate of the vertex is given by x = -b / (2a), where a = -16 and b = 144.

Plugging in the values, we get:
x = -144 / (2*(-16))
x = -144 / -32
x = 4.5

Now, to find the maximum distance, we substitute this x-value into the function:
s(4.5) = -16(4.5)^2 + 144(4.5) + 120
s(4.5) = -16(20.25) + 648 + 120
s(4.5) = -324 + 648 + 120
s(4.5) = 444

Therefore, the maximum distance above the ground is 444 feet.

B) The height of the building is the constant term in the function. In this case, it is the initial height of the object when it was projected. From the function s(t) = −16t^2 + 144t + 120, we can see that the height of the building is 120 feet.

To find the maximum distance above the ground, we need to determine the vertex of the quadratic function. The vertex can be found using the formula:

t = -b / (2a)

First, let's identify the coefficients in the equation:
a = -16
b = 144

Now we can substitute these values into the formula to find the time at which the object reaches its maximum height:

t = -144 / (2*(-16))
t = -144 / (-32)
t = 4.5

To find the maximum distance, we need to substitute this value of t back into the original equation:

s(4.5) = -16(4.5)^2 + 144(4.5) + 120
s(4.5) = -16(20.25) + 648 + 120
s(4.5) = -324 + 648 + 120
s(4.5) = 444

Therefore, the maximum distance above the ground is 444 feet (A).

To find the height of the building, we need to consider the initial height of the object when it was launched, which is represented by the constant term in the equation. In this case, the constant term is 120.

Therefore, the height of the building is 120 feet (B).

To find the maximum distance above the ground, we need to determine the vertex of the quadratic function. The vertex represents the maximum or minimum point of the function.

The formula for the vertex of a quadratic function in the form ax^2 + bx + c is given by the equation x = -b / (2a).

In our case, the quadratic function is s(t) = -16t^2 + 144t + 120, so a = -16 and b = 144.

So, the x-coordinate of the vertex, which represents the time at which the object reaches its maximum height, is:
t = -144 / (2*(-16))
t = -144 / -32
t = 4.5 seconds

To find the maximum distance above the ground, we need to substitute the value of t back into the equation s(t).

s(4.5) = -16(4.5)^2 + 144(4.5) + 120
s(4.5) = -16(20.25) + 648 + 120
s(4.5) = -324 + 768 + 120
s(4.5) = 564

Therefore, the maximum distance above the ground is 564 feet.

To find the height of the building, we need to determine the initial position of the object when it was projected upward from the top of the building.

Since s(t) represents the distance above the ground, the height of the building would be the initial position of the object, i.e., the value of s(0).

s(0) = -16(0)^2 + 144(0) + 120
s(0) = 120

Therefore, the height of the building is 120 feet.