A plane drops a hamper of medical supplies from a height of 3410 m during a practice run over the ocean. The plane's horizontal velocity was 121 m/s at the instant the hamper was dropped. What is the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean? The acceleration of gravity is 9.8 meters per second squared

Are you supposed to estimate aerodynamic drag? When dropped from that that altitude, it will definitely not be negligible.

Both the horizontal and vertical drag will be appreciable.

I would complain to the teacher

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 problem into two components: the vertical velocity and the horizontal velocity.

1. Vertical velocity:
We can start by finding the time it takes for the hamper to fall to the surface of the ocean. We can use the equation:

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

Where:
h = height (3410 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time

Rearranging the equation, we get:

t = √(2 * h / g)

Substituting the values:

t = √(2 * 3410 / 9.8) ≈ 27.70 s

Using this time, we can find the vertical velocity using the equation:

v_vertical = g * t

v_vertical = 9.8 * 27.70 ≈ 271.06 m/s (approximately)

2. Horizontal velocity:
The horizontal velocity of the hamper remains constant throughout its fall. Therefore, the horizontal velocity of the hamper is the same as the plane's horizontal velocity, which is given as 121 m/s.

3. Overall velocity:
To find the magnitude of the overall velocity, we can use the Pythagorean theorem, as the vertical and horizontal velocities are perpendicular to each other. The equation for the magnitude of the overall velocity is:

v_overall = √(v_vertical^2 + v_horizontal^2)

Substituting the values:

v_overall = √(271.06^2 + 121^2) ≈ 294.48 m/s

Therefore, the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean is approximately 294.48 m/s.