Question

A ball is dropped from a great height and the velocity of the ball as a function of time is given $v=\frac{mg}{K}(1-e^{-\left(K/m\right)t})$, where $m$ is mass of ball, $K$ is a constant and $g$ is the acceleration due to gravity. Then, at a time when $t\gg \left(m/K\right)$, the velocity of the ball becomes

• A. $v=gt$
• B. $v=\frac{mg}{K}$
• C. $v=\frac{mg}{K}t$
• D. $v=0$

Answer 1

Answer: (B)

$v=\frac{mg}{K}$

$t>>\frac{m}{K}$
$\Rightarrow e^{-\left(K/m\right)t}⟶0\Rightarrow v=\frac{mg}{K}$