Answer:
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
Answer:
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
