Question

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(sub-x)+v^2(sub-y)]^1/2

And I believe that:

v(sub-x)=v(naught-x)=v(naught)[cos (&)]
v(sub-y)=v(naught-y)-gt;
v(naught-y)=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

Answer 1

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.

Answer 2

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