A student observes an airplane fly overhead with a constant velocity parallel to

Question

A student observes an airplane fly overhead with a constant velocity parallel to the x-axis at a height of 7.60 Times 10^3 m. At time t = 0 the airplane is directly above the student, the vector P_o = 7.60 Times 10^3 j m describes the position on the airplane with respect to the student. At time t = 30.0 s, the position vector has changed to P_30 = (8.04 Times 10^3i + 7.60 Times 10^3_j) m as shown in the figure below. Ignore drag. Determine the magnitude and orientation of the airplane's position vector at t = 45.0 s. Write down a time-dependent vector P_t that describes the position of the airplane with respect to the student at any time after t = 0. Suppose the plane drops a package at the instant t = 30.0 s. What is the position vector that describes the location of the package with respect to the airplane (i.e. a vector that points from the plane's location to the package at that instant)?

Explanation / Answer

a) in time 30s, plane travels 8.04 x 10^3 m in horizontal with constant velocity.

hence v = d/t = (8.04 x 10^3) / 30 = 268 m/s

so distance traveled in 45 s will be, Dx = vt = 268 x 45 = 12060 m = 12.06 x 10^3 m

P45 = (12.06 x 10^3 i + 7.60 x 10^3 j )

b) p = (vti +P0)

p = (268t i + 7.60 x 10^3 j )

c) Initially package will have same velocity as plane.

so package will move downwards under the influence of gravity.

so r = - gt^2 /2 j

r = - 4.9t^2 j

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