# Physics

posted by .

Sorry, I asked this question earlier but I didn't clarify it well:

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.

I was told this was one dimensional however the Height (y) is equal to 5210 and the horizontal (x-axis) velocity is 133 m/s so it certainly seems to have to be 2d; also the chapter we're in is all 2 and 3 dimensional.

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

• Physics -

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