Mechanical energy of cannon ball

[physics phan]

ok,

heres the prob:

A 21.0 kg cannon ball is fired from a cannon with a muzzle speed of 1100 m/s at an angle of 37.3° with the horizontal. A second ball is fired at an angle of 90.0°. Use the law of conservation of mechanical energy to find

(b) the total mechanical energy at the maximum height for each ball. Let y = 0 at the cannon.


in (a), i had to find the height for each ball, which i did using ( mv^2/2 ) - mgh = 0

i found the first ball to be 2.27e4 m/s (using 1100sin37.3)
and the second ball to be 6.17e4

now i can find the mech energy

do you use ( mv^2/2 ) + mgh = Mech E

(i simplified changed in PE and changed in KE in the above equation)

and then add the two Mech Energies?

 

[arildno]

You will get the right answer in (a) by using (( mv^2/2 ) - mgh = 0), if v is your vertical initial speed.

But how did you reason to get this formula?

 

[physics phan]

I used the vertical velocity

in a part 1, i did 1100sin37.3 for v, and in part 2, it was a 90 degree angle

i got this formula because that's what we've been using in class... although i just posted the simplified version of it... the part where we started plugging in numbers

 

[arildno]

OK, let's go to part (b).
What does "conservation" mean?
I, at least, mean that when some quantity is conserved (in time, for example), it doesn't change it's value (in time) (Agreed?)
Therefore, a cannonball's mechanical energy does not change its value from the moment (when it is shot out from the cannon) to its "final" moment (when it has reached the top).
Was this helpful?

 

[physics phan]

OK, let's go to part (b).
What does "conservation" mean?
I, at least, mean that when some quantity is conserved (in time, for example), it doesn't change it's value (in time) (Agreed?)
Therefore, a cannonball's mechanical energy does not change its value from the moment (when it is shot out from the cannon) to its "final" moment (when it has reached the top).
Was this helpful?

I understand what you are saying but I am not quite sure what to do still...

i am confused why they told us why the cannon is at y=0... unless that is just to tell us to add two positives

also, i am not sure how to find mech energy... we went over it in class for just a couple minutes and have no clue how to derive a proper formula for this problem

 

[arildno]

Ok, here's some more:
I assume your teacher has told you that mechanical energy consists of two parts:
1. Kinetic energy
2. Potential energy
When you meet the phrase "total mechanical energy" that just means "kinetic energy + potential energy"

Why y=0?
How does this equation enable you to calculate the potential energy at the initial level, i.e. where the cannon is?

 

[physics phan]

i got it

it was 1/2 mv^2

sheeeesh x 1000


why didn't you tell me it was that easy

cmon oslo boy


keep it real dawg, aight