Answer
(a) $6.7~m/s$
(b) $-5.9~m/s$
(c) $9.6~m/s$
Work Step by Step
Let the positive direction be upward.
(a) $distance~traveled=4.0~m$, $elapsed~time=0.60~s$
$average~speed=\frac{distance}{elapsed~time}=\frac{4.0~m}{0.60~s}=6.7~m/s$
(b) $a=-g=-9.81~m/s^2$, $t=0.60~s$
$\Delta v=at$
$\Delta v=(-9.81~m/s^2)(0.60~s)=-5.886~m/s\approx-5.9~m/s$
(c) $v_{av}=\frac{v_0+v}{2}$. Since the positive direction is upward, $v_{av}=6.7~m/s$. So:
$\frac{v_0+v}{2}=6.7~m/s$. Also:
$\Delta v=v-v_0=-5.886~m/s$ -> $v=v_0-5.886~m/s$. Replace this information in the above equation:
$\frac{v_0+v}{2}=6.7~m/s$
$\frac{v_0+v_0-5.886~m/s}{2}=6.7~m/s$
$2v_0-5.886~m/s=2(6.7~m/s)$
$2v_0=13.4~m/s+5.886~m/s$
$v_0=\frac{19.286~m/s}{2}=9.643~m/s\approx9.6~m/s$
