Does this look right? Cannonball question

  • Context: Undergrad 
  • Thread starter Thread starter mathatesme
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Discussion Overview

The discussion revolves around the trajectory of a cannonball fired at a 45-degree angle with an initial velocity of 100√2 ft/sec. Participants analyze the trajectory equation, calculate the distance until it hits the ground, and determine the maximum height reached by the cannonball.

Discussion Character

  • Technical explanation, Debate/contested

Main Points Raised

  • One participant presents the trajectory equation y=x-(x/25)^2 and claims the distance until it hits the ground is 312.5 ft and the maximum height is 156.25 ft.
  • Another participant agrees with the calculations presented, indicating a consensus on the values provided.
  • Some participants express confusion regarding the calculations, specifically mentioning a potential error related to the squared term in the trajectory equation.
  • One participant acknowledges having made a similar mistake but ultimately agrees with the original claim after reviewing it.

Areas of Agreement / Disagreement

While there is some agreement on the calculations, there is also confusion and acknowledgment of potential errors, indicating that the discussion remains somewhat contested.

Contextual Notes

Participants reference specific calculations and terms, but there is uncertainty regarding the correctness of the trajectory equation and its implications for the results.

mathatesme
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Cannonball fired on 45 deg angle from horizontal. Initial position is origin. initial velocity 100sqrt(2) ft/sec. Trajectory is y=x-(x/25)^2 where y>=0.

How far till it hits the ground? 312.5

What's the max height? 156.25

Here is what I got m(a) = 2*-(1/625)a + 1 = 0 which = -(1/312)a + 312 = 0

-(1/625)*312.5^2+312.5 = 156.25

Do you agree?
 
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mathatesme said:
...
Trajectory is y=x-(x/25)^2 where y>=0.
...
Here is what I got m(a) = 2*-(1/625)a + 1 = 0 which = -(1/312)a + 312 = 0

-(1/625)*312.5^2+312.5 = 156.25

Do you agree?

Yep, I agree. :smile:
 
Last edited:
Nope... his is right. He has the [itex]25[/itex] squared too~
 
Data said:
Nope... his is right. He has the [itex]25[/itex] squared too~
Argh, my brain just blurred over this one. :smile:
 
I made that mistake too, for a second, and was about to post it when I noticed he was right :-p
 

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