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
answered it myself, thanks anyway.
To find the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean, you can use the formula you mentioned:
v = sqrt(vx^2 + vy^2)
where vx is the horizontal component of the velocity and vy is the vertical component of the velocity.
From your information, the horizontal velocity (vx) is given as 133 m/s. This means that there is no acceleration or deceleration in the horizontal direction.
For the vertical component of the velocity (vy), we need to find the initial vertical velocity (v0y). We can use the formula v0y = v0 * sin(theta), where v0 is the initial velocity and theta is the angle at which the object is dropped.
In this case, the object is dropped from a height, so the angle theta can be considered 270 degrees or -90 degrees (downward direction). Since the plane's horizontal velocity doesn't affect the angle, you can use either of these angles.
Now, the initial velocity can be calculated using the formula v0 = sqrt(vx^2 + vy^2).
Substituting the given values, we get:
v0 = sqrt((133 m/s)^2 + (0 m/s)^2) = 133 m/s
For v0y, we can use v0y = v0 * sin(theta):
v0y = 133 m/s * sin(270 degrees) = 133 m/s * (-1) = -133 m/s
Now, you can find the magnitude of the overall velocity using the formula:
v = sqrt(vx^2 + vy^2) = sqrt((133 m/s)^2 + (-133 m/s)^2) = sqrt(17689 m^2/s^2 + 17689 m^2/s^2) = sqrt(35378 m^2/s^2) ≈ 188.055 m/s
So, the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean is approximately 188.055 m/s, which is different from the given answer of 346.129 m/s.
It's possible that there was an error in your calculations or in the given answer. I would recommend double-checking your calculations and verifying the given answer with the correct solution to ensure accuracy.