Passage I $(1-7)$

Objects falling in air do not continue to accelerate indefinitely as they fall. A falling object reaches a maximum speed$($terminal speed, $v_T)$ because as it falls the upward force of air resistance increases as the object's peed relative to the air $($air speed$)$ increases. When the upward force of air resistance balances the weight of the objet, acceleration ceases. The upward force is proportional to the square of the air speed of the object and the area presented by the object perpendicular to the motion through the air. By changing from a minimum to a maximum surface area presented perpendicular to the path of the motion, a falling person can reduce his or her terminal speed by a factor of THREE.

$v_T=\sqrt[\phantom{2}]{\frac{2mg}{CA}}$

The formula gives the terminal velocity, $v_t,$ of a falling object, where $m$ is the mass of the falling object, $g$ is the acceleration due to gravity, $C$ is dimensionless and depends on the objects shape, $$ is the density of the surrounding air, and $A$ is the cross$-$sectional area of the object.

An individual with a mass of $100$ kg is released from rest from a height of $3000$ m $,$ after $700$ m below the released point it reaches the terminal speed at $50\frac{m}{s}.($Note: $g=10\frac{m}{s^2}$ and assume constant air density below $3$km$.)$