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

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

To find the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean, we can use the formula you provided:

v = √(v^2(sub-x) + v^2(sub-y))

Given:
Height of the drop, h = 5210 m
Horizontal velocity of the plane, v(naught-x) = 133 m/s
Acceleration due to gravity, g = 9.8 m/s^2

First, let's calculate the time it takes for the hamper to fall to the surface of the ocean. We can use the formula for vertical displacement:

h = v(naught-y) * t - (1/2) * g * t^2

Since the hamper is dropped vertically, v(naught-y) = 0, and we can simplify the equation:

h = - (1/2) * g * t^2

Solving for t:

t^2 = -2h / g

t = √(-2h / g)

Substituting the given value:

t = √(-2 * 5210 / 9.8)
t ≈ 32.12 s

Now, let's calculate the horizontal component of the velocity:

v(sub-x) = v(naught-x) = 133 m/s

Since the hamper is dropped vertically, the vertical component of the velocity is affected by the acceleration due to gravity:

v(naught-y) = -g * t

Substituting the values:

v(naught-y) = -(9.8 m/s^2) * (32.12 s)
v(naught-y) ≈ -314.576 m/s

Now we can calculate the magnitude of the overall velocity:

v = √(v^2(sub-x) + v^2(sub-y))

v = √((133 m/s)^2 + (-314.576 m/s)^2)
v ≈ √(17689 + 98897.81)
v ≈ √116586.81
v ≈ 341.444 m/s

Therefore, the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean is approximately 341.444 m/s, which rounds to 341.44 m/s.

To find the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean, we can break down the velocity into its horizontal and vertical components.

Given:
- Initial vertical velocity (v₀₋ᵧ) = v₀ * sin(&)
- Initial horizontal velocity (v₀₋ₓ) = v₀ * cos(&)
- Acceleration due to gravity (g) = 9.8 m/s²

Since the plane's velocity is only in the horizontal direction, it does not affect the vertical components of the hamper's motion.

The vertical component of the velocity can be calculated using the equation v₋ᵧ = v₀₋ᵧ - g * t, where t is the time of flight. In this case, since the hamper is dropped vertically downward, the time of flight is the same as the time taken to fall from the height of 5210 m. We can calculate the time using the equation s = v₀₋ᵧ * t + (1/2) * g * t², where s is the vertical distance traveled (5210 m).

By substituting the values, we have 5210 m = v₀₋ᵧ * t + (1/2) * (9.8 m/s²) * t². Solving this equation for t will give us the time of flight.

Once we have the time of flight, we can calculate the magnitude of the overall velocity using the equation v = sqrt(v₀₋ₓ² + v₋ᵧ²).

Let's calculate it step by step:

1. Calculate the time of flight:
5210 m = v₀₋ᵧ * t + (1/2) * (9.8 m/s²) * t²

Simplify the equation:
4.9 t² + v₀₋ᵧ t - 5210 = 0

Solve this quadratic equation for t. You can use the quadratic formula:
t = (-v₀₋ᵧ ± sqrt(v₀₋ᵧ² + 4 * (4.9) * (5210))) / (2 * (4.9))

The positive value of t should be used since time cannot be negative.

2. Substitute the value of t back into the equation to calculate the vertical component of the velocity (v₋ᵧ):
v₋ᵧ = v₀₋ᵧ - g * t

3. Calculate the magnitude of the overall velocity (v) using the equation v = sqrt(v₀₋ₓ² + v₋ᵧ²):
v = sqrt(v₀₋ₓ² + v₋ᵧ²)

By following these steps, you should obtain the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean, which in this case is given as 346.129 m.