Calculating Angle to Shoot Sword Over Wall for UT - 65 Characters

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Homework Help Overview

The problem involves calculating the angle at which a sword must be shot over a wall to reach a character named UT. The wall is 14 meters high, and UT is positioned 9 meters behind it. The scenario includes considerations of potential energy, projectile motion, and the heights of the individuals involved.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the energy required to get the sword over the wall and question the assumptions regarding the heights of the individuals involved. There are attempts to define variables for unknowns and explore the equations of motion relevant to projectile trajectories.

Discussion Status

The discussion is ongoing, with participants providing insights into the energy calculations and the need for additional parameters. Some suggest breaking the problem into two phases and exploring the implications of different launch angles and distances. There is no explicit consensus on a single approach, but various methods and considerations are being explored.

Contextual Notes

Participants note the lack of information regarding the height from which the sword is fired and the potential impact of the shooter’s position on the required angle. There is also mention of the need to consider horizontal motion and the effects of gravity in the calculations.

  • #31
Well, if one speed is 5.85 m/s (did the calculation again with 11.6) and the other one is 7.54 then launch speed is

√(5.58^2+7.54^2)= 9.3 m/s

The speed available from the bow is
179=(1.28 kg *v^2)/2
v=16.703

16.703 > 9.3 , meaning that it doesn't exceed the speed available from the bow and that I can use the given speed.
 
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  • #32
Superman123 said:
Well, if one speed is 5.85 m/s (did the calculation again with 11.6) and the other one is 7.54 then launch speed is

√(5.58^2+7.54^2)= 9.3 m/s

The speed available from the bow is
179=(1.28 kg *v^2)/2
v=16.703

16.703 > 9.3 , meaning that it doesn't exceed the speed available from the bow and that I can use the given speed.
Looks right.
 
  • #33
Thank you very much, I appreciate your help !:smile:
Just a very tiny question, did you use suvat-equations to construct this function for the optimal angle (π-atan(d/h))/2?
 
  • #34
Superman123 said:
Thank you very much, I appreciate your help !:smile:
Just a very tiny question, did you use suvat-equations to construct this function for the optimal angle (π-atan(d/h))/2?
Yes.
 

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