Bernoulli's Equation — Question 5

Question 5

The lowest hole does not send a jet farthest

A large open tank maintains its free surface at height $H$ above a horizontal floor. A small horizontal outlet is at height $z$ above the floor, with $0<z<H$. Use ideal Bernoulli discharge, negligible surface speed and projectile motion without air drag. The jet exits at atmospheric pressure.

1. Derive the horizontal range $R(z)$ before the jet reaches the floor.

2. Find the outlet height giving maximum range. Explain why two complementary heights $z$ and $H-z$ give the same range.

Question 5 - Solution

The lowest hole does not send a jet farthest

(a) Derivation and calculation

Bernoulli supplies the horizontal exit speed, while free fall supplies flight time:$$v=\sqrt{2g(H-z)},\qquad t=\sqrt{2z/g}.$$$$\boxed{R(z)=vt=2\sqrt{z(H-z)}}.$$$$R\to0\quad\text{as }z\to0^+\text{ or }z\to H^-.$$

(b) Result and interpretation

$$z(H-z)=H^2/4-(z-H/2)^2\le H^2/4.$$$$\boxed{z_{\mathrm{optimum}}=H/2},\qquad\boxed{R_{\max}=H}.$$$$R(H-z)=2\sqrt{(H-z)z}=R(z).$$Complementary heights interchange the speed head and flight height. A lower hole gives more speed but less flight time.

Physical check

The gravitational acceleration cancels from the ideal range because its effects on speed and fall time compensate.