## University Physics with Modern Physics (14th Edition)

(a) $a_y = \frac{5g}{4}$ (b) $a_y = \frac{3g}{4}$
The drag force $f$ is $Cv^2$, where C is a positive constant. The terminal speed $v_t$ is the speed when $Cv_t^2 = mg$. $v_t = \sqrt{\frac{mg}{C}}$ (a) Suppose that $v$ is half the terminal speed. $v = \frac{v_t}{2} = \sqrt{\frac{mg}{4C}}$ On the way up, the drag force $f$ is directed straight down. We can find the acceleration $a_y$. $\sum F_y = m~a_y$ $mg+f = m~a_y$ $mg+Cv^2 = m~a_y$ $mg+C\frac{mg}{4C} = m~a_y$ $mg + \frac{mg}{4} = m~a_y$ $a_y = \frac{5g}{4}$ (b) Suppose that $v$ is half the terminal speed. $v = \frac{v_t}{2} = \sqrt{\frac{mg}{4C}}$ On the way down, the drag force $f$ is directed straight up. We can find the acceleration $a_y$. $\sum F_y = m~a_y$ $mg-f = m~a_y$ $mg-Cv^2 = m~a_y$ $mg-C\frac{mg}{4C} = m~a_y$ $mg - \frac{mg}{4} = m~a_y$ $a_y = \frac{3g}{4}$