Fluid Mechanics: Step-by-Step Worked Examples (6)

hard • 3 min read
Tags Fluid Mechanics

Question

A large tank holds water to a depth of H=5H = 5 m. A small hole is drilled at depth h=1h = 1 m below the surface. Find the speed of water emerging from the hole and the horizontal distance from the base of the tank where the water lands. Take g=10g = 10 m/s2^2.

Solution — Step by Step

The hole is at depth h=1h = 1 m below the water surface. The exit speed is

v=2gh=2×10×1=20≈4.47 m/sv = \sqrt{2 g h} = \sqrt{2 \times 10 \times 1} = \sqrt{20} \approx 4.47 \text{ m/s}

The water depth in the tank is 5 m total. Hole is 1 m below the surface, so the hole is at height H−h=5−1=4H - h = 5 - 1 = 4 m above the base.

Water leaves the hole horizontally and falls 4 m. Using y=12gt2y = \tfrac{1}{2} g t^2:

t=2yg=2×410=0.8≈0.894 st = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 4}{10}} = \sqrt{0.8} \approx 0.894 \text{ s}

x=vt=4.47×0.894≈4 mx = v t = 4.47 \times 0.894 \approx 4 \text{ m}

Why This Works

Torricelli’s theorem v=2ghv = \sqrt{2gh} comes directly from Bernoulli’s equation applied between the water surface (at rest, atmospheric pressure) and the exit hole (also atmospheric pressure). The pressure terms cancel, the velocity at the surface is zero (large tank assumption), and we are left with v2=2ghv^2 = 2gh — the same formula as a freely falling object dropped from height hh.

Once water exits, it is in projectile motion with horizontal initial velocity vv and zero vertical velocity. Standard projectile equations give the range.

A pretty result: the range x=2h(H−h)x = 2\sqrt{h(H-h)}. This is maximum when h=H/2h = H/2 — hole exactly at the middle of the water column. JEE Advanced 2019 asked for this maximum directly.

Alternative Method

Use the range formula directly: x=2h(H−h)=21×4=2×2=4x = 2\sqrt{h(H-h)} = 2\sqrt{1 \times 4} = 2 \times 2 = 4 m. Same answer in one line — memorize this.

Students use hh as the height above the ground (not below the surface) when applying Torricelli. The depth hh in v=2ghv = \sqrt{2gh} is always measured from the water surface down to the hole. The vertical fall to the ground is a separate quantity.

Final answer: v≈4.47v \approx 4.47 m/s, range x=4x = 4 m.

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