A large steel water storage tank with a diameter of 20 m is filled with water and is open to the atmosphere (1 atm = 101 kPa) at the top of the tank. If a small hole rusts through the side of the tank, 5.0 m below the surface of the water and 20.0 m above the ground, assuming wind resistance and friction between the water and steel are not significant factors, how far from the base of the tank will the water hit the ground?

Explanation:

Their explanation is: We first need to determine the velocity of the water that comes out of the hole, using Bernoulli's equation.

P1+ρgy1 + ½ρv1² = P2+ρgy2+½ρv2²

The atmospheric pressure exerted on the surface of the water at the top of the tank and at the hole are essentially the same.

Then, P1-P2 = 0

. Additionally, since the opening at the top of the tank is so large compared to the hole on the side, the velocity of water at the top of the tank will be essentially zero. We can also set y1 as zero, simplifying the equation to:

0 = ρgy2 + ½ρv2²

Divide through by density

-gy2 = ½ v2²

y2= 5m and g =-9.81

-•-9.81×5 = ½v2²

9.81×5×2 = v2²

98.1 = v2²

v2 = √98.1

v2 = 9.9m/s

Using equation of motion to know the time of fall

We assume that the initial vertical velocity of the water is zero and the displacement of the water is -20 m

y-yo = ut + ½gt²

0-20 = 0•t -½ ×9.81 t²

-20 = -4.905t²

t² = -20÷-4.905

t² = 4.07747

t =√4.07747

t = 2.02secs

Using range formula

Then, R=Voxt

R= 9.9 × 2.02

R =19.99m

R ≈20m

The distance from the base of the tank to the ground is 20 m

Explanation:

From Bernoulli equation where P is pressure, V is velocity, ρ is density, g is acceleration due to gravity and y is the height

P₁ + 1/2ρV₁² + ρgy₁ = P₂ + 1/2ρV₂² + ρgy₂

The pressure on the surface of the water on the top of the tank and that in the hole is the same (P₁ = P₂). Since the opening at the top of the tank is large compared to the hole at the side of the tank, the velocity at the top of the tank is 0 and y₁ is 0

Therefore: 0 =  1/2ρV₂² + ρgy₂

1/2ρV₂² = -ρgy₂      g = -10 m/s²

-10(5) = -1/2V₂²

V₂ = 10 m/s

The time it would take the water to fall on the ground is given as:

d = ut + 1/2gt²

u is the initial velocity = 0 , g = -10 m/s² and the displacement (d) = - 20 m

Therefore: -20 = 1/2(10)t²

t = 2 sec

The horizontal displacement (d) can be gotten from

d = V₂t

d = 10(2) = 20 m

The distance from the base of the tank to the ground is 20 m

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s