What Is the Initial Speed of a Projectile Fired from a Cliff?

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Homework Help Overview

The problem involves a projectile fired from a cliff at an angle, with the goal of determining its initial speed based on its landing distance from the base of the cliff. The context includes elements of projectile motion, specifically focusing on horizontal and vertical components of motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss resolving the initial speed into horizontal and vertical components. There are questions about the total vertical distance and the interpretation of time in the equations used. Some participants express uncertainty about the setup and the assumptions regarding vertical displacement and time intervals.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup. Some guidance has been offered regarding the equations and the nature of the time variable, but no consensus has been reached on the overall approach or solution.

Contextual Notes

There is a mention of the vertical displacement being 20m and the potential confusion regarding the total time of flight versus the time for specific parts of the projectile's motion. The original poster has requested explanations, indicating a desire for deeper understanding.

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Homework Statement


A cannon is fired at an angle of 30 degrees above the horizontal from a cliff that is 20m above a flat river bottom. What is the initial speed of the projectile if it is found to land 40m from the base of the cliff?

Please provide me with relevant explanations as well. Thanks! :D
 
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Resolve the speed into horizontal and vertical components.
 
The speed is not provided in the question.
 
Call the speed v, then.
 
To clarify, assuming I've set this up right, if you do this ultimately you end up with two equations in v and t, which you can solve for either.
 
What I got so far was Vi cos 30t = 40
Is the total vertical distance merely 20m since the vertical displacement of the cannonball during the parabolic trajectory is 0? (Since it goes up and down the same distance?) - pardon my bad english lol!

Is the "t" in the Vi cos 30t = 40 equation the total time taken for the entire projectile motion or just the upper symmetrical part of the projectile (excluding the 20m descent)?
Thanks!
 
jysim said:
What I got so far was Vi cos 30t = 40
Is the total vertical distance merely 20m since the vertical displacement of the cannonball during the parabolic trajectory is 0? (Since it goes up and down the same distance?) - pardon my bad english lol!

Is the "t" in the Vi cos 30t = 40 equation the total time taken for the entire projectile motion or just the upper symmetrical part of the projectile (excluding the 20m descent)?
Thanks!

First part yes (don't forget to check your signs), and for the second part t is the total time (as you'd using the total time for both the horizontal and vertical components, no?)
 

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