The function h(t) = −16t2 + 60t + 95 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) = 20 + 38.7t, 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 1 through 4 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)

i need with part b

H(t) = -16t² + 60t + 95
g(t) = 20 + 38.7t

h(1) = -16(1²) + 60(1) + 95 = -16 + 60 + 95 = -16 + 155 = 139
h(2) = -16(2²) + 60(2) + 95 = -16(4) + 120 + 95 = -64 + 215 = 151
h(3) = -16(3²) + 60(3) + 95 = -16(9) + 180 + 95 = -144 + 275 = 131
h(4) = -16(4²) + 60(4) + 95 = -16(16) + 240 + 95 = -256 + 335 = 79

g(1) = 20 + 38.7(1) = 20 + 38.7 = 58.7
g(2) = 20 + 38.7(2) = 20 + 77.4 = 97.4
g(3) = 20 + 38.7(3) = 20 + 116.1 = 136.1
g(4) = 20 + 38.7(4) = 20 + 154.8 = 174.8

Between 2 and 3 seconds. 
The range of the 1st object is 151 to 131.
The range of the 2nd object is 97.4 to 136.1

h(t) = g(t) ⇒ 131 = 131

It means that the point where the 2 objects are equal is the point where the 1st object is falling down while the 2nd object is still going up.