The functions are [tex]H(t)= -16t^{2}+80t+112 [/tex] and g(t)=5.2t+20,
are both functions which give the heights of cannonballs thrown from 2 different cannons.
A.
[tex]H(3)= -16(3)^{2}+80(3)+112=208 [/tex]
[tex]H(4)= -16(4)^{2}+80(4)+112=176 [/tex]
[tex]H(5)= -16(5)^{2}+80(5)+112=112 [/tex]
[tex]H(6)= -16(6)^{2}+80(6)+112=16 [/tex]
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.