A plane is flying horizontally with speed

152 m/s at a height 2110 m above the ground,
when a package is dropped from the plane.
The acceleration of gravity is 9.8 m/s
2
.
Neglecting air resistance,A second package is thrown downward from
the plane with a vertical speed v1 = 72 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

To find 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 determine the horizontal and vertical components of the velocity separately and then combine them using vector addition.

Given information:
The plane is flying horizontally with a speed of 152 m/s.
The package is dropped vertically with no initial horizontal velocity.

First, let's consider the vertical component of the velocity. The package is thrown downward with a vertical speed v1 = 72 m/s. Since the direction is downward, the vertical velocity component is negative. So, the vertical component of the velocity is -72 m/s.

Next, let's consider the horizontal component of the velocity. Since the package is dropped vertically and there is no initial horizontal velocity, the horizontal component remains the same as the plane's horizontal velocity. So, the horizontal component of the velocity is 152 m/s.

Now, using vector addition, we can find the magnitude of the total velocity. Since the vertical and horizontal components are perpendicular to each other, we can use the Pythagorean theorem:

Total velocity magnitude = √(vertical velocity)^2 + (horizontal velocity)^2
Total velocity magnitude = √((-72 m/s)^2 + (152 m/s)^2)

Calculating the magnitude of the total velocity:

Total velocity magnitude = √(5184 m^2/s^2 + 23104 m^2/s^2)
Total velocity magnitude = √(28288 m^2/s^2)
Total velocity magnitude ≈ 168.18 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 168.18 m/s.