The function H(t) = −16t2 + 90t + 50 shows the height H(t), in feet, of a projectile after t seconds. A second object moves in the air along a path represented by g(t) = 28 + 48.8t, where g(t) is the height, in feet, of the object from the ground at time t seconds.Part A: Create a table using integers 1 through 4 for the 2 functions. Between what 2 seconds is the solution to H(t) = g(t) located? How do you know? (6 points)Part B: Explain what the solution from Part A means in the context of the problem. (4 points)
Accepted Solution
A:
Answer:Step-by-step explanation:Part A:Table for H(t) = −16t² + 90t + 50:x y1 1242 1663 1764 154Table for g(t) = 28 + 48.8t:x y1 76.82 125.63 174.44 223.2H(t) = −16t² + 90t + 50H(1) = −16(1)² + 90(1) + 50H(1) = 124H(t) = −16t² + 90t + 50H(2) = −16(2)² + 90(2) + 50H(2) = 166H(t) = −16t² + 90t + 50H(3) = −16(3)² + 90(3) + 50H(3) = 176H(t) = −16t² + 90t + 50H(4) = −16(4)² + 90(4) + 50H(4) = 154g(t) = 28 + 48.8tg(1) = 28 + 48.8(1)g(1) = 76.8g(t) = 28 + 48.8tg(2) = 28 + 48.8(2)g(2) = 125.6g(t) = 28 + 48.8tg(3) = 28 + 48.8(3)g(3) = 174.4g(t) = 28 + 48.8tg(4) = 28 + 48.8(4)g(4) = 223.2The solution to H(t) = g(t) is located between the third and fourth second. The y values of H(t) and g(t) get closer and closer relative to the time, but at the fourth second the y values are farther apart. Between these two seconds, both tables of values would have the same ordered pair.Part B: The solution from part A for H(t) and g(t) represents the point where both objects pass while flying. At this point, the two objects would collide, changing their flight paths and equations. The equations of H(t) and g(t) are no longer valid after this point.