Question

In the 2nd lab you were studying water squirting out from a hole in a cylinder filled with water into a tub. Just as a falling ball converts potential energy to kinetic energy, water pressure pushed water out of the hole by converting potential energy per unit volume (density x g x height) to kinetic energy per unit volume (density x velocity^2/2). Assume you have a 20 cm head of water above the hole and that the hole is 40 cm above ground.

How long after your release the finger holding the hole (in seconds) will it take the water to reach the ground (think mechanics)?

Answer 1

The time it takes for water to fall from a height can be determined by using the equation of motion under constant acceleration, specifically the equation:

h = v_0 * t + 0.5 * a * t^2

where h is the height, v_0 is the initial velocity (which is 0 in this case), t is the time, and a is the acceleration due to gravity. a = 9.8 m/s^2.

We can rearrange this equation to solve for t:

t = sqrt(2h/a)

Now, let's plug in the numbers for h and a:

h = 40 cm = 0.4 m
a = 9.8 m/s^2

t = sqrt(2 * 0.4 / 9.8) = sqrt(0.08163265306) = 0.2857 seconds

So, it would take approximately 0.29 seconds for the water to reach the ground after the finger holding the hole is released.