Question

describe the motion of a skydiver from the time he jumps to the time he lands safely on the ground

Answer 1

The skydiver accelerates at the beginning, while later he continues at constant velocity

Explanation:

There are two forces acting on the skydiver during its fall:

• The force of gravity, which is constant, of magnitude $mg$, where m is the mass of the skydiver and $g$ is the acceleration of gravity ($9.8 m/s^2$ downward)
• The air resistance, whose magnitude is proportional to the  speed of the skydiver: $F_r\propto v$, and which acts upward

At the beginning of its fall, the speed of the skydiver is small, therefore at the beginning the air resistance is negligible, and therefore the skydiver is accelerated downward with an acceleration of

$g=9.8 m/s^2$

However, as the skydiver falls, he gains speed; therefore, the magnitude of the air resistance increases. The net force acting on the skydiver is

$F_{net}=mg-F_r$

As the air resistance increases, the net force decreases: therefore, the acceleration of the skydiver decreases. Eventually, he will reach a speed at which the air resistance becomes equal to the force of gravity:

$mg=F_r$

when this occurs, the acceleration of the skydiver will become zero, so he will continue his fall at a constant velocity: this value of the velocity is called terminal velocity.

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