1. The problem statement, all variables and given/known data 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. 2. Relevant equations Mechanical energy = KE + UG + US (ignore spring potential energy) UG = mgh KE = (1/2)mv2 3. 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 + UG = KE + UG no UG at launch, and no KE at top of flight in the y-direction -- motion is changing direction -- so KE in the vertical direction = UG (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 + UG in the y-direction KE = (1/2)(32)(1360cos44)2 = 15313200.87 J UG = (32)(9.8)(45536.98701) = 14280399.13 J KE2 + UG2 = 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?