We are given:
The function: y = -16t² + 64
where y is the height from ground, t seconds after falling
Part A:
when the droplet would hit the ground, it's height from the ground will be 0
replacing that in the given function:
0 = -16t² + 64
16t² = 64 [adding 16t² on both sides]
t² = 4 [dividing both sides by 16]
t = 2 seconds [taking square root of both sides]
Part B:
for second droplet,
height from ground = 16 feet
time taken = t seconds
acceleration due to gravity = 10 m/s²
initial velocity = 0 m/s
h = ut + (1/2)at² [second equation of motion]
16 = (0)(t) + (1/2)(10)(t²)
16 = 5t²
t² = 16/5
t = 1.8 seconds (approx)
Therefore, the second droplet takes the least amount time to hit the ground