Answer:
The object will be 2 inches above the rest position when at approximately 0.23 seconds
Explanation:
The given parameters are;
The equation that models the distance, d, of the object in inches above or below the rest position = a×(cos(b·t))
If a = 5 and b = 5, we have;
d = 5 × (cos(5 × t))
When the object is approximately 2 inches above the ground, we have;
d = 2 = 5 × (cos(5 × t))
cos(5 × t) = 2/5
5 × t = cos⁻¹(2/5) ≈ 1.16
t = (cos⁻¹(2/5))/5 ≈ 1.16/5
t ≈ 0.23 s
The object will be 2 inches above the rest position when t ≈ 0.23 s.