Projectile motion cannon shell problem

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Homework Help Overview

The problem involves projectile motion, specifically analyzing the trajectory of a cannon shell fired horizontally from a height of 50.0 meters with an initial velocity of 60.0 m/s. The original poster attempts to determine the range of the shell's trajectory using kinematic equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the vertical and horizontal components of motion, with the original poster calculating the time of flight and range. Some participants express confidence in the calculations, while others question the formulas used and the assumptions made regarding the projectile's motion.

Discussion Status

The discussion is ongoing, with some participants offering supportive feedback on the calculations. There is a focus on confirming the correctness of the approach and the formulas applied, but no explicit consensus has been reached regarding the final outcome.

Contextual Notes

Participants note concerns about the formulas used and the assumptions related to the horizontal launch angle. There is an emphasis on understanding the underlying physics rather than simply confirming the numerical result.

Zugox
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Homework Statement


A coastal defence cannon fires a shell horizontally from the top of a 50.0m high cliff directed out to sea with a velocity of 60.0m/s.
Determine the range of the shell's trajectory.

Homework Equations


delta x = ut (horizontally)
delta y = ut + 1/2 at^2

The Attempt at a Solution


working vertically
50 = 0t + (1/2 * 9.8 * t^2)
t = 3.2

working horizontally
delta x = ut
delta x = 60 * 3.2
delta x = 192
therefore the range is 192m

I'm not too great with physics although really willing to learn and really interested in it :)
Can someone just confirm I'm right or put me on the right track
 
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I haven't done the math but your work seems right, I'm studying this area in Physics class too :D
 
MysticDude said:
I haven't done the math but your work seems right, I'm studying this area in Physics class too :D

I'm 99% sure on the math, was just worried about the formulas I used:)
 
Well, since we are firing horizontally, \theta = 0 meaning y-y_0 = \frac{-1}{2}gt^2 and you have that. Then you can just plug in for d=vt and you should get 192m.
 

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