Final answer:
The time it takes for a droplet of water to fall 0.25 m from the hole in a cylindrical tank of water, starting from rest and assuming no air resistance, is calculated using kinematic equations. The result is approximately 0.227 seconds.
Step-by-step explanation:
The question involves calculating the time it takes for a given droplet of water to fall 0.25 m from the hole in a cylindrical tank of water. This problem can be approached using the concepts of kinematics from physics. Assuming the absence of air resistance, the motion of the droplet is one of constant acceleration due to gravity (g ≈ 9.81 m/s2).
To calculate the time it takes for the water droplet to fall, we can use the kinematic equation for motion under uniform acceleration:
s = ut + 1/2 a t2
Where:
- s is the displacement (0.25 m in this case)
- u is the initial velocity (0 m/s if it starts from rest)
- a is the acceleration (9.81 m/s2 for gravity)
- t is the time (what we're solving for)
Plugging the values in, we have:
0.25 m = 0 m/s × t + 1/2 × 9.81 m/s2 × t2
We can simplify this to:
0.25 m = 4.905 m/s2 × t2
Now, solving for t gives us:
t2 = 0.25 m / 4.905 m/s2
t = sqrt(0.25 m / 4.905 m/s2)
t ≈ 0.227 s
Therefore, it takes approximately 0.227 seconds for the droplet to fall 0.25 m from the hole.