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 .75 m/s leaves the end of a horizontal channel at the top of a vertical wall h=2.35 m high and falls into a pool. a) how far from the wall will the water land? will that space behind the waterfall be wide enough for a pedestrian to walk through? b) To sell her plan to the city council, the architect wants to build a model to standard scale, 1/12 actual size. How fast should the water flow in the channel of the model? Using x=1/2gt^2 i found t to equal .69 seconds how do i plug that back in correctly?

Explanation / Answer

A landscape architect is planning an artificial waterfall in a city park. Water flowing at 1.72 m/s will leave the end of a horizontal channel at the top of a vertical wall 2.80 m high, and from there the water falls into a pool.
To sell her plan to the city council, the architect wants to build a model to standard scale, one-thirteenth actual size. How fast should the water flow in the channel in the model?

You want to maintain a 1/10 horizontal displacement with the model.

time it takes to fall the distance
h1 = 1/2*g*t1^2
t1 = sqrt(2*h1/g)

The horizontal distance, s1 = t1*1.72 m/s

Your new horizontal distance for the model must be 1/10 of that horizontal distance, i.e. s2 = t1*0.172 m

Your new height is 1/0 the original, so h2 = 0.28 m.

The time it would take to fall
t2 = sqrt(2*h2/g)

The rate needs to be s2/t2 = t1*0.172/t2
rate = 0.172 * sqrt(2*h1/g) / sqrt(2*h2/g)

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