An airline passenger drops a coin while the plane is moving horizontally at 240 m/s. What is the velocity of the coin relative to the plane and relative to the Earth when it strikes the floor 1 m below its point of release?

###### Solution.

The plane is an inertial frame of reference and an observer (passenger) in the plane sees the coin fall straight down.

The final velocity can be found using the equation:

v_{f}^{2} = v_{i}^{2} – 2g(y_{f}-y_{i})

Substituting known values into the equation, we get the velocity of the coin relative to the plane

v_{f}^{2} = 0^{2} – 2*9.8*(0-1)

v_{f} ≈ – 4.43 m/s

Note, we choose the negative root because we know that the velocity is directed downwards.

Now, an observer on the Earth sees the coin move almost horizontally.

The horizontal motion and the vertical motion are independent, the final vertical velocity for the coin relative to the Earth is v_{y} = − 4.43 m/s, the same as the velocity of the coin relative to the plane.

The horizontal component of the velocity of the coin equals the velocity of the plane v_{x} = 240 m/s.

The x and y components of velocity can be combined to find the magnitude of the final velocity:

v = √(v_{x}^{2} + v_{y}^{2})

Thereby,

v = √(240^{2} + (-4.43)^{2}) ≈ 240.04 m/s

The direction is given by:

θ = tan^{-1}(v_{y}/v_{x})

hence,

θ = tan^{-1}(- 4.43/240) = -1.06^{0}