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**1. A landscape architect is planning an artificial water fall in a city park. Water flowing at 1.56 m/s will leave the end of a horizontal channel at the top of a vertical wall 2.01 m high and from there fall into a pool.**

(a) How wide a space will this leave for a walkway at the foot of the wall under the waterfall?

(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 in the channel flow in the model?

(a) How wide a space will this leave for a walkway at the foot of the wall under the waterfall?

(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 in the channel flow in the model?

**2. s=v(initial)*t + 1/2*at^2**

**3. Okay, so I was able to get part A by calculating horizontal and vertical motion.**

vertical motion:

s=v(initial)*t + 1/2*at^2

-2.01=1/2*(-9.8m/s^2)*t^2

t^2= .410

t=.64 seconds

then, I plugged the time into this equation for horizontal displacement:

s=(1.56m/s)t + 0*t^2

s=(1.56m/s)(.64s)

s=.999m=1m

Part B is what I cannot figure out.

I tried dividing the initial velocity of 1.56m/s by 12, but I got the wrong answer of .13m/s.

If you can please give me a hint about how I'm supposed to solve for part B, I'd really appreciate it! :D

vertical motion:

s=v(initial)*t + 1/2*at^2

-2.01=1/2*(-9.8m/s^2)*t^2

t^2= .410

t=.64 seconds

then, I plugged the time into this equation for horizontal displacement:

s=(1.56m/s)t + 0*t^2

s=(1.56m/s)(.64s)

s=.999m=1m

Part B is what I cannot figure out.

I tried dividing the initial velocity of 1.56m/s by 12, but I got the wrong answer of .13m/s.

If you can please give me a hint about how I'm supposed to solve for part B, I'd really appreciate it! :D

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