Consider a water jar of radius R that has water filled up to height H and is kept on a stand of height h. Through a hole of radius r (r << R) at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is x, then:

At the hole (height $$h$$): Velocity = $$v_1$$, Radius = $$r$$, Area = $$\pi r^2$$.

At the ground (height 0): Velocity = $$v_2$$, Radius = $$x$$, Area = $$\pi x^2$$.

$$v_1 = \sqrt{2gH}$$

Using $$v^2 = u^2 + 2as$$: $$v_2^2 = v_1^2 + 2gh$$

$$v_2^2 = 2gH + 2gh = 2g(H + h)$$

$$v_2 = \sqrt{2g(H + h)}$$

Using equation of continuity, $$A_1 v_1 = A_2 v_2$$, 

$$\pi r^2 v_1 = \pi x^2 v_2$$

$$x^2 = r^2 \left( \frac{v_1}{v_2} \right)$$

$$x^2 = r^2 \frac{\sqrt{2gH}}{\sqrt{2g(H + h)}}$$

$$x^2 = r^2 \sqrt{\frac{H}{H + h}} = r^2 \left( \frac{H}{H + h} \right)^{1/2}$$

$$x = r \left( \frac{H}{H + h} \right)^{1/4}$$