Response:
0.14 s
Detailed breakdown:
s = -2.7 t² + 40t + 6.5
Set s = 12
12 = -2.7t² + 40t + 6.5 Rearranging yields
-2.7t² + 40t + 6.5 - 12 = 0
-2.7t² + 40t - 5.5 = 0
Utilize the quadratic formula

a = -2.7; b = 40; c = -5.5




x = 7.41 ± 7.27
x₁ = 0.14; x₂ = 14.68
The graph indicates roots at x₁ = 0.134 and x₂ = 14.68.
The surface of the Moon stands at -12 ft. Thus, the ball will reach a height of 12 ft above the Moon’s surface (crossing the x-axis) at 0.14 s.
The second root indicates when the ball is again 12 ft above the lunar surface as it descends.