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therest
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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 + UG + US (ignore spring potential energy)
UG = 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 + 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?