Answer
(a) $0.31~s$
(b) $0.11~m$
Work Step by Step
(a) Remember: the motion on the way up is symmetrical to the motion on the way down. (See conceptual question 16)
So, assuming the positive direction be upward with origin at the level of the syringe, when the liquid returns to the level of the syringe $v=-v_0$.
$v_0=1.5~m/s$, $v=-1.5~m/s$, $a=-g=-9.81~m/s^2$
$v=v_0+at$
$-1.5~m/s=1.5~m/s+(-9.81~m/s^2)t$
$(9.81~m/s^2)t=3.0~m/s$
$t=\frac{3.0~m/s}{9.81~m/s^2}=0.31~s$
(b) When the liquid reaches the maximum height its velocity is zero.
$v_0=1.5~m/s$, $v=0$, $a=-g=-9.81~m/s^2$
$v^2=v_0^2+2a\Delta x$
$0^2=(1.5~m/s)^2+2(-9.81~m/s^2)\Delta x$
$(19.62~m/s^2)\Delta x=2.25~m^2/s^2$
$\Delta x=\frac{2.25~m^2/s^2}{19.62~m/s^2}=0.11~m$
