The function H(t) = −16t^2 + 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)
*** I think it is bewteen 3 and 4 because those are the two closest***
Part B: Explain what the solution from Part A means in the context of the problem. (4 points)
***they will meet after between 3 or 4***
Part A: To find the solution to H(t) = g(t), we need to create a table using integers 1 through 4 for the two functions.
First, let's evaluate the function H(t) = -16t^2 + 90t + 50 for values of t from 1 to 4.
For t = 1:
H(1) = -16(1)^2 + 90(1) + 50 = -16 + 90 + 50 = 124
For t = 2:
H(2) = -16(2)^2 + 90(2) + 50 = -64 + 180 + 50 = 166
For t = 3:
H(3) = -16(3)^2 + 90(3) + 50 = -144 + 270 + 50 = 176
For t = 4:
H(4) = -16(4)^2 + 90(4) + 50 = -256 + 360 + 50 = 154
Next, let's evaluate the function g(t) = 28 + 48.8t for values of t from 1 to 4.
For t = 1:
g(1) = 28 + 48.8(1) = 28 + 48.8 = 76.8
For t = 2:
g(2) = 28 + 48.8(2) = 28 + 97.6 = 125.6
For t = 3:
g(3) = 28 + 48.8(3) = 28 + 146.4 = 174.4
For t = 4:
g(4) = 28 + 48.8(4) = 28 + 195.2 = 223.2
Comparing H(t) and g(t) values, at t = 3, H(t) = 176 and g(t) = 174.4, and at t = 4, H(t) = 154 and g(t) = 223.2. Therefore, the solution to H(t) = g(t) is between 3 and 4 seconds.
Part B: The solution to H(t) = g(t) represents the time at which the two objects, described by the functions H(t) and g(t), are at the same height. In the context of the problem, it means that after approximately 3 to 4 seconds, the projectile described by H(t) and the second object moving along the path g(t) will be at the same height. This could be the point at which they intersect or cross each other.