Answer
(a) $2.3~s$
(b) $20~m/s$
Work Step by Step
$v_0=-4.2~m/s$, $x_0=35~m$, $a=-g=-9.81~m/s^2$
(a) $x=0$
$x=x_0+v_0t+\frac{1}{2}at^2$
$0=x_0+v_0t-\frac{g}{2}t^2$
$t=\frac{-v_0±\sqrt {v_0^2-4(\frac{-g}{2})x_0}}{2(-\frac{g}{2})}=\frac{-v_0±\sqrt {v_0^2+2gx_0}}{-g}$
$t=\frac{-(-4.2~m/s)±\sqrt {(-4.2~m/s)^2+2(9.81~m/s^2)(35~m)}}{-9.81~m/s^2}$
$t_1=-3.1~s$ (not valid)
$t_2=2.3~s$
(b) $x=15~m$
$\Delta x=x-x_0=15~m-35~m=-20~m$
$v^2=v_0^2+2a\Delta x$
$v=±\sqrt {(-4.2~m/s)^2+2(-9.81~~m/s^2)(-20~m)}$
$v=±20~m/s$
$v=-20~m/s$ because the bag is moving downward.
$speed=|v|=20~m/s$
