Answer
(a) $32~m$
(b) $25~m/s$
Work Step by Step
$v_0=6.5~m/s$, $a=-g=-9.81~m/s^2$, $v=0$.
$v^2=v_0^2+2a\Delta x$
$0=(6.5~m/s)^2+2(-9.81~m/s^2)\Delta x$
$(19.62~m/s^2)\Delta x=(6.5~m/s)^2$
$\Delta x=\frac{(6.5~m/s)^2}{19.62~m/s^2}=2~m$
That is, $\Delta x=2~m$, no matter what the value of $x_0$
$\Delta x=2~m$
$x-x_0=2~m$
$x=x_0+2~m$
If $x_0=20~m$, then $x=22~m$. If $x_0=30~m$, then $x=32~m$
(b) $v_0=6.5~m/s$, $a=-g=-9.81~m/s^2$, $x_0=30~m$, $x=0$.
$\Delta x=x-x_0=0-30~m=-30~m$
$v^2=v_0^2+2a\Delta x$
$v=±\sqrt {(6.5~m/s)^2+2(-9.81~m/s^2)(-30~m)}$
$v=±25~m/s$
$v=-25~m/s$ because the bag is moving downward before it lands.
$speed=|v|=25~m/s$
