Using projectiles to solve for 2-D motion problems

• ea1369
In summary, the landscape architect plans to build a 2.01 m high artificial water fall in a city park with water flowing at 1.56 m/s. To accommodate a walkway at the foot of the wall, the space will be 1 m wide. To create a model for the city council, the water in the channel should flow at a rate of 0.13 m/s, as calculated by rescaling the vertical displacement and finding the new time for free-fall.
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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?

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

Last edited:
For part B you need to rescale the vertical displacement to match the model and find the new time for free-fall. The velocity can be found from the equation for the rescaled horizontal displacement.

fzero said:
For part B you need to rescale the vertical displacement to match the model and find the new time for free-fall. The velocity can be found from the equation for the rescaled horizontal displacement.

thanks so much! I just went back and figured out the answer:)

1. How do you determine the initial velocity of a projectile in 2-D motion?

The initial velocity of a projectile in 2-D motion can be determined by breaking it into its horizontal and vertical components. The horizontal component is the initial velocity in the x-direction, and the vertical component is the initial velocity in the y-direction. These can be calculated using the equations v0x = v0cosθ and v0y = v0sinθ, where v0 is the initial velocity and θ is the angle of projection.

2. How do you find the time of flight for a projectile in 2-D motion?

The time of flight for a projectile in 2-D motion can be calculated using the equation t = 2v0y/g, where v0y is the initial velocity in the y-direction and g is the acceleration due to gravity. This formula assumes no air resistance and a level surface.

3. How do you determine the maximum height reached by a projectile in 2-D motion?

The maximum height reached by a projectile in 2-D motion can be calculated using the equation h = (v0y)2/2g, where v0y is the initial velocity in the y-direction and g is the acceleration due to gravity. This formula assumes no air resistance and a level surface.

4. What is the range of a projectile in 2-D motion?

The range of a projectile in 2-D motion can be calculated using the equation R = v0xt, where v0x is the initial velocity in the x-direction and t is the time of flight. This formula assumes no air resistance and a level surface.

5. How do you account for air resistance in 2-D motion problems?

To account for air resistance in 2-D motion problems, you can use the formula FD = ½ρAv2, where FD is the force of air resistance, ρ is the air density, A is the cross-sectional area of the projectile, and v is the velocity of the projectile. This force can then be used in equations of motion to adjust for air resistance.

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