What is the optimal angle to set a catapult for attacking a medieval city?

Click For Summary

Homework Help Overview

The problem involves determining the optimal angle for launching projectiles from a catapult to clear the walls of a medieval city, which has square walls measuring 550 m in length and 16 m in height. The launching point is 100 m away from the wall, and the initial speed of the projectiles is 80 m/s.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to calculate angles that allow the projectile to clear the wall while considering the distance from the wall and the height of the wall. Some mention using kinematics and equations of motion to analyze the trajectory.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup and the necessary calculations. Some have suggested focusing on the maximum height needed to clear the wall and the range required to ensure the projectile lands beyond the wall.

Contextual Notes

There is some ambiguity regarding the dimensions and layout of the walls, as well as the exact requirements for the projectile's trajectory. Participants express varying levels of understanding and readiness to engage with the problem.

Ravey
Messages
5
Reaction score
0
1. A medieval city has the shape of a square and is protected by walls with length 550 m and height 16 m. You are the commander of an attacking army and the closest you can get to the wall is 100 m. Your plan is to set fire to the city by catapulting heated rocks over the wall (with an initial speed of 80 m/s). At what range of angles should you tell your men to set the catapult? (Assume the path of the rocks is perpendicular to the wall. Round the answer to one decimal place. g ≈ 9.8 m/s2)



2. Nada clue other than some kinematics and vectors



3. Tried using kinematics but failed miserably. Help! C'mon it's about medieval stuff! ;)
 
Physics news on Phys.org
I guess that means the walls are 550m on each side. Could be total length I suppose but that isn't clear. Anyway you have the distance from the wall

(see attachment for picture)

So I guess you aim it high enough so it just clears the the top of the first wall on the way down, and low enough so it arches over the first wall to the foot of the second wall -- if you can. So set up your equations of motion and get cracking. :-)
 

Attachments

  • catapault.jpg
    catapault.jpg
    4.6 KB · Views: 565
Last edited:
Your minimum angle would be where max height of the trajectory = 16m, because your projectile has to get over the wall. Maximum would be range = 100m to throw the projectile past the wall.
 
Ya LCKurtz is right, you don't seem to now where to start so...

Your at 100 m of the first wall and 650 m of the second you now the highest altitude will be half-way 325m and... It's your homework. :-p