Question

A hole is punched in the side of a 29.8 cm tall container, full of water. If the water is to shoot as far as possible horizontally, how far from the bottom of the container should the hole be punched? Answer in units of cm.

Neglecting friction losses, how far from the side of the container will the water land?
Answer in units of cm.

Answer 1

The distance the from the bottom of the container at which the hole

should be depends on pressure head of the water.

Response:

• The location from the to punch the hole is 14.9 cm.

• The furthest horizontal distance from the side of the container the water will land is 29.8 cm.

How can the distance from the bottom of the hole should be punched be calculated?

First part:

The distance the water will shoot out horizontally is given from the formula which is derived ;

$H - h = \mathbf{\dfrac{1}{2} \cdot g \cdot t^2}$

$t = \mathbf{\sqrt{\dfrac{2 \cdot (H-h)}{g} }}$

$Horizontal \ velocity, \ v = \mathbf{\sqrt{2 \cdot g \cdot h}}$

Which gives;

Where;

H = The height of the cylinder

h =

$Horizontal \ velocity, \ v = \sqrt{2 \cdot g \cdot h}$

$Horizontal \ distance, \ x = v \cdot t = \sqrt{2 \cdot g \cdot h} \cdot t = \mathbf{\sqrt{2 \cdot g \cdot h} \cdot \sqrt{\dfrac{2 \cdot (H-h)}{g} }}$

$Horizontal \ distance, \ x = \sqrt{2 \cdot g \cdot h} \cdot \sqrt{\dfrac{2 \cdot (H-h)}{g} }= \mathbf{2 \sqrt{h \cdot (H - h)}}$

x² = 4·h·H - 4·h²

$Maximum \ x^2 \ is \ at \ h = -\dfrac{4 \cdot H}{2 \cdot (-4)} = \dfrac{H}{2}$

$\dfrac{H}{2}$ = $\dfrac{29.8 \, cm}{2}$ =  14.9 cm. from the top is $\dfrac{H}{2}$ = 14.9 cm from the bottom

Therefore;

If the water is to shoot as far as possible, the distance from the the

bottom of the container, at which the hole should be is punched is half

way from the bottom of the container which is 14.9 cm. from the bottom.

Second part:

The horizontal distance from the side of the container the water will land is therefore;

$x_{max} = 2 \times \sqrt{14.9 \times (29.8 - 14.9)} = 2 \times \sqrt{14.9^2} = \mathbf{29.8}$

• The (furthest) distance from the water will land from the side of the container is $x_{max}$ = 29.8 cm.

Learn more about Newton's law of motion here:

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