A jet of water squirts out horizontally from a hole near the bottom of the tank,
Assume that y = 1.19 m and x = 0.544 m. What is the speed of the water coming out of the hole?
d = vt - horizontal distance = 0.544
s = 1/2 at^2 - vertical distance = 1.19
so,
t = √(2s/a)
v = d/t = d/√(2s/a) = .544/√(2*1.19/9.8) = 1.10m/s
To find the speed of the water coming out of the hole, we can use the principle of conservation of energy.
First, let's denote the potential energy of the water at the surface of the tank as PE₁, and the potential energy of the water at the hole as PE₂. Since both points are at the same height, the change in potential energy is zero: PE₁ - PE₂ = 0.
The potential energy of an object is given by the formula PE = m * g * h, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
Since the water jet is horizontal, it means there is no change in height (h = 0). Therefore, the potential energy at the hole becomes PE₂ = m * g * 0 = 0.
Now, let's consider the kinetic energy of the water at the hole. The kinetic energy of an object is given by the formula KE = (1/2) * m * v², where v is the speed of the object.
Since the water is leaving the tank, there is no external force doing work on it, meaning there is no energy change. As a result, the kinetic energy at the surface of the tank is equal to the kinetic energy at the hole: KE₁ = KE₂.
Substituting the equation for kinetic energy, we get:
(1/2) * m₁ * v₁² = (1/2) * m₂ * v₂²,
where m₁ and m₂ are the masses of the water at the surface of the tank and at the hole, respectively, and v₁ and v₂ are their respective speeds.
Since the water is incompressible and the density of water is constant, we can assume that m₁ = m₂ = m (mass of water).
After canceling out the common factors and rearranging the equation, we find:
v₂ = sqrt(v₁² + 2 * g * h),
where v₂ is the speed of the water coming out of the hole, v₁ is the speed of the water at the surface, g is the acceleration due to gravity, and h is the depth of the water surface below the hole.
Given that y = 1.19 m (depth of the water surface) and x = 0.544 m (horizontal distance from the hole), we can use Pythagoras theorem to find h:
h = sqrt(y² - x²) = sqrt(1.19² - 0.544²) = sqrt(1.4161 - 0.2969) = sqrt(1.1192) = 1.057 m.
Now, we can substitute the values into the equation to find v₂:
v₂ = sqrt(v₁² + 2 * g * h) = sqrt(0 + 2 * 9.8 * 1.057) = sqrt(20.5928) = 4.54 m/s.
Therefore, the speed of the water coming out of the hole is approximately 4.54 m/s.