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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?

User Njlarsson
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Answer:

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

User Sahas
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