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