Using projectiles to solve for 2-D motion problems

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SUMMARY

The discussion focuses on solving a 2-D motion problem involving projectiles, specifically in the context of an artificial waterfall design. The landscape architect needs to calculate the horizontal distance for a walkway under a waterfall and the scaled velocity for a model. Using the equation for vertical motion, the time of free-fall was determined to be 0.64 seconds, resulting in a horizontal displacement of 1 meter. For the model, the correct approach involves rescaling both vertical displacement and time to find the appropriate water flow velocity.

PREREQUISITES
  • Understanding of kinematic equations, specifically s = v(initial)*t + 1/2*at^2
  • Knowledge of projectile motion principles in two dimensions
  • Ability to perform unit conversions and scaling for model representation
  • Familiarity with free-fall motion and gravitational acceleration (9.8 m/s²)
NEXT STEPS
  • Learn how to apply kinematic equations to real-world projectile motion problems
  • Research scaling techniques for physical models in engineering design
  • Study the effects of gravity on projectile motion in different environments
  • Explore advanced projectile motion simulations using software tools like MATLAB or Python
USEFUL FOR

Students in physics, landscape architects, engineers, and anyone involved in designing models that require understanding of projectile motion and scaling principles.

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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:
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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:)
 

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