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?

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

20

step-by-step explanation:

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6