A tin can is filled with water to a depth of 30 cm. A hole 11 cm above the bottom of the can produces a stream of water that is directed at an angle of 34° above the horizontal.

Question

(a) Find the range of this stream of water.
(b) Find the maximum height of this stream of water.

Answer 1

Use the Bernoulli equation to get the velocity of the water leaving the hole. There is 0.19 m of water above the hole.
The water pressure there is (rho)*g*h above ambient pressure, which is
1000 kg/m^3*9.81m/s^2*0.19 = 1864 N
(rho is the density of water)

The water flows through the hole at a velocity given by
(1/2)*rho*V^2 = 1864 N
V = 1.93 m/s

Now consider the water in air as a ballistics problem. Drops are launched with a vertical velocity component of 1.93 sin 34 = 1.08 m/s and a horizontal component of 1.60 m/s. The water hits the ground 0.11 m below the hole. Compute the time it takes to hit the ground and multiply that by 1.60 m/s to get the range of the stream.

The maximum height can be computed by computing that the stream reaches maximum height at (1.08 m/s)/g = 0.11 s after leaving the hole. Multiply that by the average vertical velocity component as it reaches maximum height, 0.54 m/s. The stream rises 5.6 cm above the hole, or 16.6 cm above the base.