The function h(t) = −16t2 + 80t + 112 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) = 20 + 5.2t, 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 3 through 6 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)

The functions are and g(t)=5.2t+20,

are both functions which give the heights of cannonballs thrown from 2 different cannons.

A.





g(3)=5.2(3)+20=31.6
g(4)=5.2(4)+20=40.8
g(5)=5.2(5)+20=46
g(6)=5.2(6)+20=51.2

up to the 5th second, the ball from H is higher than the one from g, 
then from the 6th second the ball from H is lower than the ball from g,

this means that the solution of H(t)=g(t), is between the 5th and 6th seconds.

B. The solution of part A means that during the 3rd, until it falls, the ball from H is falling to the ground, but it is still higher than the ball g which is rising, until a point between the 5th and 6 th seconds, where the balls are at the same height for just one instant, then the ball H continues falling, while g rises up for a few more second.