An object is thrown up from the top of a building that is 480 feet high. The equation y=-16t^2+208+480 gives the object’s height (in feet) above the ground at any time t (in seconds) after the object is thrown.

a) y(0 s) = 480 ft, y(1 s) = 672 ft, y(2 s) = 832 ft, y(3 s) = 960 ft, y(4 s) = 1056 ft, y(5 s) = 1120 ft, y(6 s) = 1152 ft, b) See attachment included below, c) The part of the graph that represent the physical phenomenon corresponds to the interval such that  and , d) The intercepts represent the values of t such that y is equal to zero. Physically speaking, it represents the instant when the object hits the ground and one intercept is reasonable due to the fact that time is a positive variable. ()

Step-by-step explanation:

The complete statement is: "An object is thrown up from the top of a building that is 480 feet high. The equation gives the object's height (in feet) above the ground at any time t (in seconds) after the object is thrown.

a) What is the height of the object after 0, 1, 2, 3, 4, 5 and 6 seconds?

b) Sketch the graph of the equation .

c) What part of the graph represents the physical aspects of the problem?

d) What are the intercepts of the graph? What do they mean?"

a) The height of the object at each instant is calculated by evaluating the function:

t = 0 s

t = 1 s

t = 2 s

t = 3 s

t = 4 s

t = 5 s

t = 6 s

b) The graph of this function is attached below as image.

c) The part of the graph that represent the physical phenomenon corresponds to the interval such that  and .

d) The intercepts represent the values of t such that y is equal to zero. Physically speaking, it represents the instant when the object hits the ground and one intercept is reasonable due to the fact that time is a positive variable. ()

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