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202211231154 Solution to 1969-HL-PHY-I-2

An aeroplane flying with a uniform speed v in a small circle of radius r at a constant altitude h releases a body. Assuming that air resistance can be neglected, find the angle of inclination between the ground and the line of sight of the pilot when he observes the body touching the ground.

Roughwork.

The angular speed \omega of the aeroplane is

\displaystyle{\omega =\frac{v}{r}}

and its position in Cartesian coordinate system

(x,y,z)=(r\cos (\omega t),r\sin (\omega t),h).

WLOG assume the top view be

so that

s=\sqrt{(vt-r\sin \omega t)^2+(r-r\cos \omega t)^2}.

Since the altitude h is related to the time of flight t by

\displaystyle{h =\frac{1}{2}gt^2},

the angle of inclination \alpha is in such

\displaystyle{\tan\alpha =\frac{h}{s}}

as to be desired.