What is the altitude of the sun at noon on december 22, as seen from a place on the tropic of cancer?

108.217 °

Explanation:

Day of year = 356 = d (Considering year of 365 days)

Latitude of Tropic of Cancer = 23.5 °N

Declination angle

δ = 23.45×sin[(360/365)(d+284)]

⇒δ = 23.45×sin[(360/365)(356+284)]

⇒δ = 5.2832 °

Altitude angle at solar noon

90+Latitude-Declination angle

= 90+23.5-5.2832

= 108.217 °

∴ Altitude angle of the Sun as seen from the tropic of cancer on December 22 is 108.217 °

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air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.