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Carnegie Mellon University



Problem 54 Hard Difficulty

A landscape architect is planning an artificial waterfall in a city park. Water flowing at 0.750 $\mathrm{m} / \mathrm{s}$ leaves the end of a horizontal channel at the top of a vertical wall $h=2.35 \mathrm{m}$ high and falls into a pool (Fig. P3.54). (a) How far from the wall will the water land? Will the space behind the waterfall be wide enough for a pedesbe wide enough for a pedestrian walkway? (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 flow in the channel in the model?


(a) $d = 0.52 \mathrm { m }$ . Yes
(b) $v _ { 0 x } ^ { \prime } = 6.28 \mathrm { mg } ^ { - 1 }$


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Video Transcript

So let's first try toe derive an equation for time we can say that doubts of why is equaling the why initial t plus 1/2 gt squared. We know that the initial velocity is going to be zero and in this case we can say that here. Negative h choosing downwards to be negative or upwards to be positive. Negative h is gonna be equaling 1/2 times negative g times T square solving for time. This would be equal in the square root of two times the height divided by G. So this would be the time we know that the horizontal displacement is equaling the axe initial multiplied by time. And so the horizontal displacement. We can simply say X would be equaling the ex initial multiplied by the square root of two H over G and we can solve. This would be equaling 0.750 meters per second, multiplied by the square root of two times 2.35 meters divided by 9.8 meters per second squared. And we find that the exposition or the horizontal um range is equaling 0.519 meters and so we can say that the horizontal distance of the water land point from the vertical wall is essentially 0.519 meters and well, the space behind the waterfall be but wide enough for a pedestrian walkway. We can say no so we can say no. Ah, the horizontal range of waterfall is small, so the space behind the waterfall is not sufficient for a pedestrian walkway. So no, So no pedestrian walkway again are horizontal ranges 0.519 meters. This would be your answer for part a four part B. Then we can say that uh, the actual size of the linear dimensions is reduced to 1/12 of its initial value. So the new height of the wall H crime is equaling H over 12 and the new horizontal distance is gonna be ex crime, and this would be X over 12. And so we can say that then ex prime, this equaling the ex initial prime multiplied by the square root of two H crime over G and weak. And then, um, software v ex initial crime. This would be equaling ex prime over the square root of two h crime over G, and we can then substitute in these values and say that this would be equaling X over 12. Divided by this would be the square root of two times H over 12 all divided by G once again and so v X initial prime is equaling X over 12 multiplied by six g over h and we can solve. So this would be 0.519 meters divided by 12 multiplied by six times 9.8 meters per second squared divided by 2.35 meters, and we find that then the new initial X velocity is 0.217 meters per second. So we can say that the required projected velocity of the water flow through the channel in the new model 0.217 meters for a second. That would be our final answer to part B. That is the end of the solution. Thank you for watching

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