Question

A landscape architect is planning an artificial waterfall in a city park. Water flowing at 1.67 m/s will leave the end of a horizontal channel at the top of a vertical wall h = 3.70 m high, and from there the water falls into a pool.(a) How wide a space will this leave for a walkway at the foot of the wall under the waterfall? (b) To sell her plan to the city council, the architect wants to build a model to standard scale, one-twelfth actual size. How fast should the water in the channel flow in the model?

Answer 1

Answer:

Step-by-step explanation:

vertical component of the velocity vy = 0 m/s

horizontal component of the velocity vx = 1.67 m/s

t, time for the water to reach the ground = √(2h / g) derived from the equation h = ut + 1/2 gt²

t = √ ( 2 × 3.70 / 9.8) = 0.869 s

a) the space left will = vx × t = 1.67 m/s ×   0.869 s = 1.45 m

b) re scaling the height from which the water traveled = 3.70 m / 12 = 0.308 m

then the time for the travel in the model = √(2h/g) = √ ( 2×0.308 / 9.8) = 0.251 s

then the horizontal displacement re scaling = 1.45 m / 12 = 0.121 m

0.121 m = vx × 0.251 s

how fast should the water in the channel flow vx  = 0.121 m / 0.251 s = 0.482 m/s

Answer 2

Using a model, an architect or designer are able to test and demonstrate ideas

The values for the width left for the walkway and the speed of the water in the model are;

(a) The width of the space the water falling will leave for a walkway is approximately 1.45 m

(b) The speed of the water in the channel flow in the model is approximately 0.48 m/s

The reason the above values are correct are given as follows:

The known parameters are:

The rate at which the water is flowing, vₓ = 1.67 m/s

The height of the vertical wall, h = 3.70 m

(a) Required:

• The width of the walkway at the foot of the wall under the water fall

Solution:

The width of the walkway at the foot of the wall is given by the horizontal distance the water travels before reaching the ground

The time it takes the water to reach the ground, t, is given as follows;

h = (1/2)·g·t²

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ t = √(2·h/g)

The width of the walkway, w, is equal to the horizontal velocity of the water flowing and the time it takes the water to reach the foot of the wall

∴ w = vₓ × t, which gives; w = vₓ × √(2·h/g)

Plugging in the values gives;

w = 1.67 m/s × (2 × 3.70 m/9.81 m/s²) ≈ 1.45 m

The width of the space it will leave for a  walkway at the foot of the wall under the waterfall, w ≈ 1.45 m

(b) The known parameters:

The scale of the standard scale model the architect wants to build = (1/12) actual size

Required:

• The speed with which the water will flow in the channel in the model

Solution:

The height of the wall in the model, $h_m$ = h/12

∴ $h_m$ = 3.70/12

The time it takes the water to reach the foot of the wall, $t_m$, is given as follows;  $t_m$ = √(2·$\mathbf{h_m}$/g)

The width of the space for a walkway in the model, $w_m$, is given as follows;

$w_m$ = w/12

From w = vₓ × √(2·h/g) by substitution gives; $w_m$ = $v_{xm}$ × √(2·$h_m$/g)

$\mathbf{v_{xm}}$ = $w_m$/(√(2·$h_m$/g)) = (w/12)/(√(2 × h/12/g))

∴ $v_{xm}$ = (1.45/12)/(√(2 × (3.70/12)/9.81)) ≈ 0.48

The speed of the water in the channel, $v_{xm}$ ≈ 0.48 m/s

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