a. time to fall:
h=1/2 g t^2
t= sqrt(2h/g)=you do it
horizontal distance: timetoFall*1.70
b. t=time above * sqrt(1/12)
horizontal distance=1/12*origwideth*v
solve for v.
you did not give the original sidewalk width.
a landscape architect is planning an artificial waterfall in a city park. water flowing at 1.70 m/s will leave the end of a horizontal channel at the top of a vertical wall h =2.35 m high . and from there it will fall into a pool ( Fig . P4.18) (a) will the space behind the waterfall be 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, which is one -twelfth actual size . how fast should the water flow in the channel in the model ?
2 answers
Plz,I wanna know this anwer