A cylindrical wooden reservoir [height=8m, diameter=4m] used to collect rainwate

Question

A cylindrical wooden reservoir [height=8m, diameter=4m] used to collect rainwater is attached to an elevated platform [height=2m]. A bored hunter fires a bullet that makes a hole 5cm from the bottom of the reservoir. The resulting stream of water lands in a pail located some distance from the edge of the reservoir.
a) If the pail is 3m from the edge of the reservoir, how far above the hole was the surface of the water?
b) If the hole has a radius of 1.5cm, how fast (in meters per second) will the surface of the water drop?

Explanation / Answer

I understand what this is saying. You have water coming out from the side of the barrel, 5 cm from the bottom. You should use Torricelli's theorem to solve this problem (http://en.wikipedia.org/wiki/Torricelli%27s_law). The problem makes us assume that the water is gushing out of the reservoir, going 3m over and 2.05m down.

How long does it take an object to drop 2.05m?
Using the familiar equation for position and acceleration
x=1/2 at^2
2.05m=.5 (9.81m/s^2)t^2
t=sqrt(.417)=.646s
That is the amount of time it takes something to drop 2.05m according to gravity.
In the same amount of time, it needs to travel 3m.
Using x/t=v, 3m/.646s=4.64m/s
According to Torrcelli's theorem

v=sqrt(2g(y2 - y1))

where g is gravity

y2 is the height of the water

y1 is the height of the hole

v is the velocity of the water coming out of the hole

4.64m/s=sqrt(2(9.81m/s^2)(y2-.05m))

21.53m^2/s^2=(19.6m/s^2)(y2-.05m)

y2=21.53/19.6 +.05m = 1.149m

Therefore the location of the top of the water is 1.15m above the hole or 1.20m from the bottom of the pail.

The velocity of the water coming out of the hole is 4.64m/s and the hole has a radius of 1.5cm.

What I'm going to do here is say the area of the hole times the speed of the water flowing through it is a measure of the volume of water going through it per second. This rate will of course change as the water goes down but they don't care about that so neither will I.

Area x velocity = Volume/second

(.015m)^2 pi x 4.64m/s = .0033(m^3 / s)

This change of volume/second is also true for the surface of the water.

Areaxvelocity =.0033(m^3/s)

(2m)^2 pi x height/s = .0033(m^3/s)

height/s = .00258m/s

That is the initial rate of the change in height of the top of the reservoir.

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