Question

Water stands at a depth H in a tank whose side wails are vertical. A hole is made on one of the walls at a depth h below the water surface. Find at what distance from the foot of the wall does the emerging stream of water strike the floor?

Answer 1

x = 2h
`\sqrt{(H)/(h) -1 }`

Step-by-step explanation:

This is an exercise in fluid mechanics, let's find the speed of the water in the hole

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

where the subscript 1 is for the tank surface and the subscript 2 is for the depth of the hole

the pressure inside and outside the tank is the same

P₁ = P₂

we must measure the distance from the same reference point, let's locate it on the surface of the water

y₁ = 0

y₂ = h

Suppose the gap is small compared to the diameter of the tank

v₁ «v₂

v₂² = 2 g (0-h)

This is the speed of the outlet water in the tank, as the force is horizontal this speed is horizontal.

Let's use the projectile launch ratios

vₓ =
`√(2g |h|)`

y = y₀ + v₀ t - ½ g t²

the height when reaching the floor is y = 0,

the initial height is measured from the floor therefore y₀ = H-h

0 = (H-h) + 0 - ½ g t²

t =
`\sqrt{(2(H-h))/(g) }`

we look for the distance x traveled

x = vₓ t

x =
`√(2g |h| ) \ √( 2(H-h)/g)`

x =
`√(4 h (H-h))`

x = 2h
`\sqrt{(H)/(h) -1 }`