A landscape architect is planning an artificial waterfall in a city park. Water

Question

A landscape architect is planning an artificial waterfall in a city park. Water flowing at 1.60 m/s will leave the end of a horizontal channel at the top of a vertical wall h = 3.75 m high, and from there the water falls into a pool (see figure). (a) Will the space behind the waterfall be wide enough tor a pedestrian walkway? (Assume that the average pedestrian walkway is 1 m wide.) Yes No (b) To sell her plan to the city council, the architect wants to build a model to a scale, which is one-seventeenth actual size. How fast should the water flow in the channel in the model? 0.28 Your response differs from the correct answer by more than 10%. Double check your calculations. m/s A landscape architect is planning an artificial waterfall in a city park. Water flowing at 1.60 m/s will leave the end of a horizontal channel at the top of a vertical wall h = 3.75 m high, and from there the water falls into a pool (see figure). (a) Will the space behind the waterfall be wide enough for a pedestrian walkway? (Assume that the average pedestrian walkway 1 m wide.) Yes No (b) To sell her plan to the city council, the architect wants to build a model to a scale, which is one-seventeenth actual size. How fast should the water flow in the channel in the model? 0.53 Your response differs from the correct answer by more than 10%. Double check your calculations. m/s A landscape architect is planning an artificial waterfall in a city park. Water flowing at 1.60 m/s will leave the end of a horizontal channel at the top of a vertical wall h = 3.75 m high, and from there the water falls into a pool (see figure). (a) Will the space behind the waterfall be wide enough for a pedestrian walkway? (Assume that the average pedestrian walkway is 1 m wide.)

Explanation / Answer

Y = Uyt - 0.5gt^2

-3.75 = 0 -0.5*9.8*t^2

t = 0.87 s

X = Ux*t = 1.6*0.87 = 1.4 m

So, Yes the pedestrian will safely walk away

B)if it is modelled to the scale

h = 3.75 / 17 = 0.22 m

x = 1/17 = 0.058 m

y = uyt - 0.5gt^2

0.22 = 0 - 0.5*9.8*t^2

t = 0.211s

X = ux*t

0.058 = ux*0.211

ux = 0.27 m/s

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