The function H(t) = −16t2 + 96t + 80 shows the height H(t), in feet, of a projectile after t seconds. A second object moves in the air along a path represented by g(t) = 31 + 32.2t, 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 2 through 5 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 have that

H(t) = −16t² + 96t + 80
g(t) = 31 + 32.2t

Part A)
see the attached table
the solution to H(t) = g(t) is between t=4 sec and t=5 sec
for t=4 sec
H(4)=208 ft and g(4)=159.8 ft
so
H(t) > g(t)

for t=5 sec
H(5)=160 ft and g(5)=192 ft
so
H(t) < g(t)

that change of H(t) from being greater to becoming smaller tells me that the solution is in that interval [4, 5]

Part B)Explain what the solution from Part A means in the context of the problem
when H(t)=g(t) means that the projectile destroyed the object N 2 in the interval [4,5]

to find the exact solution

−16t² + 96t + 80=31 + 32.2t
−16t² + 96t + 80-31-32.2t=0
−16t² + 63.8t + 49=0

using a graph tool
see the attached figure

the exact solution is
t=4.65 sec