Physics
posted by John on .
A plane drops a hamper of medical supplies from a height of 5210 m during a practice run over the ocean. The plane’s horizontal velocity was 133 m/s at the instant the hamper was dropped.
The acceleration of gravity is 9.8 m/s^2.
What is the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean?
I have a formula for this:
v=[v^2(subx)+v^2(suby)]^1/2
And I believe that:
v(subx)=v(naughtx)=v(naught)[cos (&)]
v(suby)=v(naughty)gt;
v(naughty)=v(naught)[sin (&)]
where '&' is the angle at which the object is dropped(theta if you will). the problem is I can't get the right answer; I think maybe I'm not using the correct angle, but the problem seems to imply that the angle is 270 degrees (or 90 degrees if you prefer) since it's a straight drop downward. I don't know if the plane's velocity affects the angle '&', but I didn't think it would.
The answer is given as: 346.129m

The plane is horizontal, no angle.
What is the hampers initial KE?
1/2 m*133^2
What is the hampers intial PE?
m*g*5210
What is the hampers final KE?
1/2 m 133^2+ m*g*5210
so final velocity can be found...
1/2 m vf^2=above; m divides out, solve for Vf
another way:
Vf^2=Vi^2+2gh is a more direct way. 
I'm sorry, but what do PE and KE mean?
Also, I'm not sure what your equations are saying; what is 1/2 m?
Also, not to argue the point of horizontal vs. 2d, but the chapter we're in is about motion in 2d or 3d so I'm not sure why there would be a 1d question.
the url for the question is:
webphysics.iupui.edu/152/Hw/HW04.pdf
question #4 if that helps