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)

We are given a function H(t) that represents the height of a cannon ball after t seconds as:

and a second cannon ball is represented by the function g(t) as:

PART A:

  t           0            1             2            3

H(t)        12          44          44           12

g(t)         15.2       25.2      40.4       55.6

Hence, between t= 2 seconds and t=3 seconds the ball will meet

     such that H(t)=g(t)

Since, we know that the Height H(t) decreases from 44 feet to 12 feet between t=2 to t=3 seconds.

and height g(t) increases from 40.4 feet to 55.6 feet between t=2 to t=3 seconds.

Hence, the two cannon balls will definitely meet between t=2 to t=3 seconds.

and the time at which they meet is calculated by solving:

As t can't be negative.

Hence, we get:

t=2.109 seconds

PART B:

The solution from PART A means that the one of the ball first reach the highest point at 44 feet and then returns back to the initial position and hence follows a parabolic path while the second cannon ball reach a greater height with the increase in time and hence in this phenomena the two balls will definitely meet.

mx+89b(-98)nb

step-by-step explanation:

f(x)=4

step-by-step explanation: