The function h(t) = −16t2 + 48t + 12 shows the height h(t), in feet, of a cannon ball after t seconds. a second cannon ball moves in the air along a path represented by g(t) = 10 + 15.2t, 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 0 through 3 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)

Part A:
To determine the values of the times to which the height of the two cannon balls are the same, we equate the given functions.
    H(t) = g(t)
Substitute the expressions for each.
  -16t² + 48t + 12 = 10 + 15.2t

Transpose all the terms to the left-hand side of the equation.
 -16t² + (48 - 15.2)t + (12 - 10) = 0

Simplifying,
   -16t² + 32.8t + 2 = 0

The values of t from the equation are 2.11 seconds and -0.059 seconds

Part B:
In the context of the problem, only 2.11 seconds is acceptable. This is because the second value of t which is equal to -0.059 seconds is not possible since there is no negative value for time.