Answer:
A) between seconds 1 and 2
B) one ball reaches a maximum and falls, the other one always rises
Step-by-step explanation:
H(t) = -16t² + 32t + 50
g(t) = 30 + 30.4t
A) To complete the table replace the values of t in the function, for example: H(2) =-16(2)² + 32(2) + 50 = 50
t H(t) g(t)
0 50 30
1 66 60.4
2 50 90.8
3 2 121.2
The solution to H(t) = g(t) is located between seconds 1 and 2, because before t = 1, H(t) > g(t), and after t = 2 g(t) > H(t)
B) The ball that follows function H(t) increase its height, reaches a maximum and, then, decreases its height. The ball that follows function g(t) increases its height all the time. In the beginning, the ball that follows function H(t) increases its height faster than the other ball, but after it reaches its maximum height, its height starts to decrease, giving the opportunity to the other ball to reach it.