A plane is flying horizontally with speed 251 m/s at a height 2180 m above the ground, a package is thrown downward from the plane with a vertical speed v1 = 80 m/s.

What is the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground?
Answer in units of m/s

sqrt (251^2 + 80^2)

To get the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground, we need to calculate the resultant velocity vector by considering both the horizontal and vertical components.

Given:
Horizontal speed of the plane (v_plane) = 251 m/s
Height of the plane (h_plane) = 2180 m
Vertical speed of the package (v1) = 80 m/s

Since the package is thrown downwards, its initial vertical velocity would be negative, i.e., v1 = -80 m/s.

Now, let's calculate the horizontal component of the velocity. Since the plane is flying horizontally, the horizontal velocity of the package will be the same as the horizontal velocity of the plane.

Horizontal component of the velocity (v_horizontal) = v_plane = 251 m/s

To calculate the vertical component of the velocity, we need to consider the initial vertical velocity (v1) and the acceleration due to gravity (g = 9.8 m/s²).

Vertical component of the velocity (v_vertical) = v1 + g * t
Since the package is just thrown downwards, its time of flight (t) will be 0.

v_vertical = -80 m/s + (9.8 m/s² * 0)
v_vertical = -80 m/s

Now, we can calculate the magnitude of the total velocity using the Pythagorean theorem:

Magnitude of the total velocity = √(v_horizontal² + v_vertical²)
Magnitude of the total velocity = √((251 m/s)² + (-80 m/s)²)
Magnitude of the total velocity = √(63001 m²/s² + 6400 m²/s²)
Magnitude of the total velocity = √69401 m²/s²
Magnitude of the total velocity ≈ 263.53 m/s

Therefore, the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground is approximately 263.53 m/s.