Man shot out of cannon question

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically analyzing the trajectory of a man shot out of a cannon. Participants are tasked with determining the distance traveled and the velocity upon impact with a wall, considering various parameters such as time of flight, launch angle, and initial velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of distance using the x-component of velocity and question the validity of using certain equations for final velocity. There is also exploration of the y-component of velocity and its implications on the overall motion.

Discussion Status

Some participants have offered guidance on calculating both x and y components of velocity, while others express uncertainty about the height of the buildings and the implications for displacement. Multiple interpretations of the problem are being explored, with no explicit consensus reached.

Contextual Notes

Participants note the ambiguity regarding the heights of the buildings and the initial conditions of the projectile motion, which may affect the calculations. There are references to specific values and equations, but the discussion remains open-ended without definitive resolutions.

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http://img263.imageshack.us/my.php?image=helpxy5.jpg
(pic attached)

Questions:
How far did man travel?
What was velocity upon hitting the wall? ( is it sufficent to cause death)

Known:
Time from cannon shot --> impact = 14 seconds ( on pic it says 16, but its actually 14)
Angle of cannon = 36.86 degrees
velocty = 30m/s

Homework Equations



vf^2 = vi^2+ viT ?

The Attempt at a Solution



i know that to find the distance traveled, is just the x-component.

THerefore distance = 30cos(36.86)*14 = 335.42 feet?

To find final velocity, I would think to use vf^2 = vi^2+ viT, but this is only for linear motion.

Please help!
 
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kdog3682 said:
http://img263.imageshack.us/my.php?image=helpxy5.jpg
(pic attached)

Questions:
How far did man travel?
What was velocity upon hitting the wall? ( is it sufficent to cause death)

Known:
Time from cannon shot --> impact = 14 seconds ( on pic it says 16, but its actually 14)
Angle of cannon = 36.86 degrees
velocty = 30m/s

Homework Equations



vf^2 = vi^2+ viT ?

The Attempt at a Solution



i know that to find the distance traveled, is just the x-component.

THerefore distance = 30cos(36.86)*14 = 335.42 feet?

To find final velocity, I would think to use vf^2 = vi^2+ viT, but this is only for linear motion.

Please help!
when the man is going to land on the other building, he is going to land at angle thus he will have both X and Y components to his velocity. Since there is no force (thus no acceleration) in the X direction, the X component of his velocity will remain the same. Thus you must calculate the Y component of his velocity as well. Are the buildings the same height - thus what was his displacement in the Y direction? WHat is the acceleration i tneh Y direction only? What was his initial velocity? For how long was he in the air? Find the final velocity in the Y direction from this.

now you can calculate his final velocity using both X and Y components
 
https://www.physicsforums.com/newreply.php?do=newreply&noquote=1&p=1345358[/b]

"Thus you must calculate the Y component of his velocity as well.
Are the buildings the same height - thus what was his displacement in the Y direction? WHat is the acceleration i tneh Y direction only? What was his initial velocity? For how long was he in the air? Find the final velocity in the Y direction from this.

now you can calculate his final velocity using both X and Y components"

You do not know if the buildings are the same heights. There is also the chance, that he lands on top of the building.
the Y displacement is 30sin(36.86)14 + 3 (the 3 is from the canon's height) - 4.9(14^2) = -693.88 feet.

A = -9.81?
Vi intial = 30 m/s or 0?
air time = 14s
using equation, vi^2 = 30^2 + 2(-9.81)*335 ( 335 is delta x) which equals
-5672.7... which is not the right answer. can sombody verify the right answers T_T
 
Last edited by a moderator:
alright if you don't know the displacement, its fine

you still know the initial velocity, the acceleration, the time of flight and thus you can find the final velocity

you don't need to find the displacement to find the final velocity...
 

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