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 ?

Plz,I wanna know this anwer

To determine if the space behind the waterfall will be wide enough for a pedestrian walkway, we need to calculate the range of the water as it falls from the top of the vertical wall. We can use the equations of motion to solve for the range.

Let's consider the vertical motion of the water. The initial velocity (Vi) in the vertical direction is 0 m/s since the water is flowing horizontally. The height (h) of the wall is 2.35 m.

Using the equation for vertical motion:

h = Vi * t + (1/2) * g * t^2,

where g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight.

At the top of the wall, the water has zero vertical displacement, so the above equation becomes:

0 = 0 * t + (1/2) * 9.8 * t^2,

0 = (4.9) * t^2.

Solving for t, we find:

t = 0 or t = √(0/4.9) = 0.

This implies that the time of flight is zero, indicating that the water falls vertically and directly into the pool without any horizontal displacement. Therefore, there will be no space behind the waterfall for a pedestrian walkway.

Moving on to part (b), we need to calculate the required flow rate in the model to maintain the same proportions as the actual waterfall. Since the model is one-twelfth of the actual size, the flow rate should be scaled proportionately.

The flow rate of water (Q) is given by:

Q = A * v,

where A is the cross-sectional area of the channel and v is the velocity of the water flow.

Let's assume that the cross-sectional area of the channel remains the same in the model as in reality.

Since the model is one-twelfth the size, the velocity in the model (v_model) can be calculated as:

v_model = (1/12) * v_actual.

Therefore, the water should flow at a speed one-twelfth the actual velocity in the model in order to maintain the correct proportion.

To determine if the space behind the waterfall will be wide enough for a pedestrian walkway, we need to consider the trajectory of the falling water.

(a) The water leaves the end of the horizontal channel with a velocity of 1.70 m/s and then falls vertically into a pool. The height of the vertical wall, h, is given as 2.35 m. To determine the width of the space needed behind the waterfall, we can use the formula for the horizontal range of a projectile:

R = (v^2 * sin(2θ)) / g

Where:
R is the range (horizontal distance traveled by the water)
v is the initial velocity of the water (1.70 m/s)
θ is the angle of projection (which in this case is 90 degrees for a vertical fall)
g is the acceleration due to gravity (9.8 m/s^2)

Plugging in the given values, we have:

R = (1.70^2 * sin(2 * 90)) / 9.8

R = 2.89 m

So the space needed behind the waterfall should be at least 2.89 meters wide to accommodate a pedestrian walkway.

(b) Now, let's calculate the required speed for the water flow in the model, given that the architect wants the model to be one-twelfth the actual size. Since the model is in a smaller scale, the speed of water flow needs to be adjusted accordingly.

If the desired speed for the actual waterfall is 1.70 m/s, we need to find the corresponding speed for the model. We can assume that all other factors (such as the height of the wall) remain the same.

Using the scale factor of 1/12, the speed of water flow in the model can be calculated as:

Model speed = Actual speed / Scale factor

Model speed = 1.70 m/s / 12

Model speed ≈ 0.14 m/s

Therefore, the water should flow at approximately 0.14 m/s in the channel of the model for it to accurately represent the actual waterfall.

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.