Kinetic and potential energy in firing a cannon

[timnswede]

Homework Statement

A 20.0 kg cannonball is fired from a cannon with a muzzle speed of 1000 m/s at an angle of 37.0° with the horizontal. A second ball is fired at an angle of 90.0°.

(a) Use the isolated system model to find the maximum height reached by each ball.

(b) What is the total mechanical energy of the ball-Earth system at the maximum height for each ball? Let y = 0 at the cannon.

Homework Equations

0=ΔKsys+ΔUsys
Emech=Usys+Ksys

The Attempt at a Solution

I understand how to do part a). The ball fired straight up has initial kinetic energy of (½)(20(1000^2)=(20)(9.8)h and h=51020m. Same idea for the second ball and h=18479m. I'm really confused about part b) though, apparently the total Mechanical energy for each ball is equal to the initial kinetic energy, which is (½)(20)(1000^2). But for the ball fired straight up I thought it would be Emech=-(½)(20)(1000^2)+(20)(9.8)(51020), which is wrong. I don't really understand how to find Emech.

 

[NascentOxygen]

Perhaps they want you to consider the recoil/rebound of the Earth? The cannon balls hurtles away in one direction, the Earthmoves away in another. Both carry K.E.

Does the textbook give you the answer?

 

[timnswede]

Yes, and that is the answer I found when I looked it up online. This is the answer on the guide for the problem:

 

[NascentOxygen]

So they are saying the mechanical energy of the ball remains constant throughout its flight, and mechanical energy is the sum of kinetic and gravitational P.E.

 

[timnswede]

I got the right answer if I assume there to be no initial kinetic or potential energy at the cannon. So for the one at the angle the initial kinetic would be (.5)(20)(1000cos37^2) + mgh. And the one straight up would just have potential at the height. I'm not sure if that is the correct way of thinking of it, and it is also pretty much the opposite of what the guide has.

 

[NascentOxygen]

The total energy of both balls is identical. They each begin their flight with the same muzzle velocity, some of it transforms into P.E., but there are no losses. The vertically-directed ball at some stage converts all of its K.E. into gravitational P.E. The ball fired at an inclination converts only some of its K.E. into P.E.

 

[Areil]

i still don't understand why the initial kinetic energy is equal to the total mec energy since tot mech energy = delta k + delta u, and delta e would be equal to u2-k1.
but here the answer is delta E= only k1

 

[haruspex]

tot mech energy = delta k + delta u,

No, the total change in mechanical energy (from just after firing) would be that, and would be zero. The total mech energy does not change: it is what you denote as k1 immediately after firing (PE being zero by choice of y=0).