Answer:
The water droplet is in the air for two seconds.
Step-by-step explanation:
You can solve this by finding the times at which its height is zero. The lesser of the two values will be upon launch, and the greater upon landing:
y = -16x² + 32x
Let y = 0
-16x² + 32x = 0
factor -16 out:
x² - 2x = 0
complete the square:
x² - 2x + 1 = 1
and solve for x
(x - 1)² = 1
x - 1 = ±1
x = 1 ± 1
x = 0, 2
So the water is at a height of zero at zero seconds and two seconds. It's thus in the air for two seconds.
