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***
You want to find:
−16t^2 + 90t + 50 = 28 + 48.8t
16t^2 -41.2t - 22 = 0
by the formula:
t = (41.2 ± √3105.44)/32
= appr 3.03 or -.454
so you are correct, it is between 3 and 4, much closer to 3 seconds
Part A: To find the solution to the equation H(t) = g(t), we need to create a table using integers 1 through 4 for both functions.
First, let's calculate the values for H(t) using the given function:
H(1) = -16(1)^2 + 90(1) + 50 = -16 + 90 + 50 = 124
H(2) = -16(2)^2 + 90(2) + 50 = -64 + 180 + 50 = 166
H(3) = -16(3)^2 + 90(3) + 50 = -144 + 270 + 50 = 176
H(4) = -16(4)^2 + 90(4) + 50 = -256 + 360 + 50 = 154
Next, let's calculate the values for g(t) using the given function:
g(1) = 28 + 48.8(1) = 28 + 48.8 = 76.8
g(2) = 28 + 48.8(2) = 28 + 97.6 = 125.6
g(3) = 28 + 48.8(3) = 28 + 146.4 = 174.4
g(4) = 28 + 48.8(4) = 28 + 195.2 = 223.2
Now, let's compare the values in the table.
H(t): 124 | 166 | 176 | 154
g(t): 76.8 | 125.6 | 174.4 | 223.2
We can see that H(t) and g(t) intersect between 3 and 4 seconds. This means that there is a common point where both objects have the same height. However, the closest integer values in the table are H(3) = 176 and g(4) = 223.2. So, the solution to H(t) = g(t) is located between the 3rd and 4th seconds.
Part B: The solution found from Part A, which is that H(t) = g(t) is located between the 3rd and 4th seconds, means that at some point in time between these two seconds, both objects will have the same height. In the context of the problem, the first object represents a projectile and the second object represents another object moving along a different path in the air. The two objects will meet or intersect at a certain height between the 3rd and 4th seconds. This information is useful if we want to know when the two objects will have the same vertical position or height in the air.
Part A: To create a table for the two functions, H(t) and g(t), using integers 1 through 4, we substitute the values of t into the equations and calculate the corresponding heights.
For H(t):
- H(1) = -16(1)^2 + 90(1) + 50 = 124 feet
- H(2) = -16(2)^2 + 90(2) + 50 = 170 feet
- H(3) = -16(3)^2 + 90(3) + 50 = 158 feet
- H(4) = -16(4)^2 + 90(4) + 50 = 90 feet
For g(t):
- g(1) = 28 + 48.8(1) = 76.8 feet
- g(2) = 28 + 48.8(2) = 125.6 feet
- g(3) = 28 + 48.8(3) = 174.4 feet
- g(4) = 28 + 48.8(4) = 223.2 feet
To find between what two seconds the solution to H(t) = g(t) is located, we check the values of H(t) and g(t) at each second.
From the table, we can see that the heights of both objects intersect or become equal between 3 and 4 seconds. At 3 seconds, H(t) is approximately 158 feet, and g(t) is approximately 174.4 feet. At 4 seconds, H(t) is approximately 90 feet, and g(t) is approximately 223.2 feet. Since the heights become equal between these two time intervals, we conclude that the solution to H(t) = g(t) is located between 3 and 4 seconds.
Part B: The solution from Part A means that the two objects, represented by the functions H(t) and g(t), will meet or have the same height between 3 and 4 seconds. In the context of the problem, it suggests that both projectiles will be at the same height at some point during this time interval.