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 1.009 m/s will leave the end of a horizontal channel at the top of a vertical wall h = 3.90 m high and falls into a pool. (a) How far from the wall will the water land? 9 m Will the space behind the waterfall be wide enough for a pedestrian walkway? (Assume the water must land 2 m from the wall to provide adequate space for a person to walk beneath the waterfall.) Yes (b) To sell her plan to the city council, the architect wants to build a model to standard scale, one-sixteenth actual size. How fast should the water flow in the channel in the model? IM HAVING TROUBLE WITH B, CAN YOU PLEASE GIVE ME THE EXPLANATION IN MY VALUES!

Explanation / Answer

Here is what I solved before. Please modify the figures as per your question. Let me know if you want some clarification on it. Please rate 5 stars if I succeeded in helping you. 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? 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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