Answer:
2.385 m/s
Step-by-step explanation:
Here there is a transformation of potential energy of the water at the top to Kinectic energy at the hole
This can be expressed as
Potential energy= Kinect energy
mgh= 1/2m v²
Where m= mass of water
h= Lenght below the level of the water that is in the tank.
v= the speed of the water as it leaves the hole
g= acceleration due to gravity= 9.81m/s
Here there is 'm' at both sides which can cancelled beach other out.
Then we have
gh= ¹/₂v²
making v² subject of the formular
v²=gh/(¹/₂)
v=√(2gh)
V=√2g h
V= √(2×9.81×0.29)
V=2.385 m/s
Therefore, the speed of the water as it leaves the hole is 2.385 m/s