Does this look right? Cannonball question

  • Context: Undergrad 
  • Thread starter Thread starter mathatesme
  • Start date Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
mathatesme
Messages
4
Reaction score
0
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?
 
Physics news on Phys.org
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