Answer Mech Energy of 32kg Cannon Ball at Max Height

[therest]

Homework Statement

A 32 kg cannon ball is fired from a cannon with muzzle speed of 1360 m/s at an angle of 44◦ with the horizontal. The acceleration of gravity is 9.8 m/s² . What is the total mechanical energy at the maximum height of the ball? Answer in units of J.

Homework Equations

Mechanical energy = KE + U_G + U_S (ignore spring potential energy)
U_G = mgh
KE = (1/2)mv²

The Attempt at a Solution

I actually thought I understood the concepts behind this one fine, but I keep getting the wrong answer.

First I found the height:
Theoretically, mechanical energy is conserved, so E at launch = E at top of flight (w/ max height)
KE + U_G = KE + U_G
no U_G at launch, and no KE at top of flight in the y-direction -- motion is changing direction -- so
KE in the vertical direction = U_G
(1/2)(32 kg)(1360sin44 m/s) = (32 kg)(9.8 m/s²)h
h = 45536.98701 m

Then, I attempted to find the mechanical energy at the top of the flight.
Known facts:
* the mass has a vertical acceleration and a horizontal velocity of 1360cos44 m/s
* the mass has both kinetic energy (it is moving) and potential gravitational energy.
* I guessed that the total mechanical energy would be the resultant of these two using the Pythagorean theorem since the two energies are perpendicular to each other.

E at top = KE in the x-direction + U_G in the y-direction
KE = (1/2)(32)(1360cos44)² = 15313200.87 J
U_G = (32)(9.8)(45536.98701) = 14280399.13 J

KE² + U_G² = resultant²
resultant = 20938574.93 J

But... that's the wrong answer.
Do the directions of the vectors not matter? Can someone provide an explanation for this?

 

[ehild]

Energy is scalar, has no direction, you simply have to add the different kinds of energy.

"KE² + UG² = resultant²" this is wrong.

As for your method, you overcomplicate the calculations. Conservation of the mechanical energy means that the total energy E= KE + PE at the muzzle is the same as at the maximum height.

ehild