Shooting a cannonball over a cliff

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Homework Help Overview

The problem involves a cannon firing a projectile towards a cliff, with specific parameters including the distance from the cannon to the cliff and the angle of projection. The goal is to determine how far the projectile lands past the edge of the cliff, considering its trajectory and the height of the cliff.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to analyze the projectile's motion, questioning the validity of their calculations and the interpretation of the problem statement.
  • Some participants suggest that the two solutions from the quadratic equation represent the times at which the projectile reaches the height of the cliff on its way up and down.
  • Others propose that the horizontal distance should be calculated based on the longer time value to determine where the projectile lands relative to the cliff.
  • There is a consideration of whether the cannon is positioned at the bottom of a valley, which affects the interpretation of the problem.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the implications of their calculations. Some have provided insights into the projectile's behavior at different points in its trajectory, while others are questioning the initial setup and parameters of the problem.

Contextual Notes

There are indications of potential misunderstandings regarding the problem's setup, particularly concerning the relative positions of the cannon and the cliff. Participants are also grappling with the implications of their calculations and the requirement for a specific answer from the online homework system.

clarineterr
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Homework Statement



A cannon 60m horizontally from the base of a 25m cliff is fired with a speed of 32.6m/s at an angle of 43 degrees above the horizontal towards the cliff. How far does the shell land past the edge of the cliff assuming its elevation is constant

Homework Equations



y = v0yt -.5gt^2

x= v0xt

The Attempt at a Solution



I plug in the to the first equation

25m = 32.6sin(43)t -4.9t^2

and the quadratic formula gives me two answers: 2.46s and 2.06s

so I plug this into the second equation giving me distances of 49.0m and 58.6

which seems to mean that the cannonball does not actually clear the cliff...but my online how system is asking for an answer.
 
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clarineterr said:

Homework Statement



A cannon 60m horizontally from the base of a 25m cliff is fired with a speed of 32.6m/s at an angle of 43 degrees above the horizontal towards the cliff. How far does the shell land past the edge of the cliff assuming its elevation is constant

Homework Equations



y = v0yt -.5gt^2

x= v0xt

The Attempt at a Solution



I plug in the to the first equation

25m = 32.6sin(43)t -4.9t^2

and the quadratic formula gives me two answers: 2.46s and 2.06s

so I plug this into the second equation giving me distances of 49.0m and 58.6

which seems to mean that the cannonball does not actually clear the cliff...but my online how system is asking for an answer.

If it gives 2 answers, isn't one when it passes the height going up and the other when it lands?. So don't you first need to figure what the horizontal velocity is? Then take the longer time and figure the horizontal distance from the gun (2.46*32.6*cos(43)) then subtract the distance from the base of the cliff?

Edit: I see. Your solutions are not reaching the cliff. Check your quadratic again or maybe enter -1.4 m then
 
I get 2.44 sec and 2.09 sec. Which falls even a little shorter.

Are you sure you have the right statement of the problem?

As stated, I would have to wonder.
 
clarineterr said:

Homework Statement



A cannon 60m horizontally from the base of a 25m cliff is fired with a speed of 32.6m/s at an angle of 43 degrees above the horizontal towards the cliff. How far does the shell land past the edge of the cliff assuming its elevation is constant

The way I read this, the cannon is sitting at the bottom of a valley shooting a cannon ball towards a cliff that is 60m away and 25m high.

Therefore, shouldn't you be looking to see where the canon is vertically at t where xhorizontal = 60m?

I get about 24.887m at t = 2.51655 for xvertical. Assuming this is just a rounding error, the shell would land 0m past the cliff's edge.
 
tkahn6 said:
Therefore, shouldn't you be looking to see where the canon is vertically at t where xhorizontal = 60m?

That's going to happen at two points; one on the way up and one on the way back down. You found the first.
 
negitron said:
That's going to happen at two points; one on the way up and one on the way back down. You found the first.

In other words, it's true that the cannon ball just barely avoids hitting the cliff, but it's still moving after scraping by and will eventually hit the ground.
 

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