Question

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)

Answer 1

The functions are $H(t)= -16t^{2}+80t+112$ and g(t)=5.2t+20,

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

A.
$H(3)= -16(3)^{2}+80(3)+112=208$
$H(4)= -16(4)^{2}+80(4)+112=176$
$H(5)= -16(5)^{2}+80(5)+112=112$
$H(6)= -16(6)^{2}+80(6)+112=16$

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.