Answer
(a) $1.0~m$ below the stalactite or $3.0~m$ above the poll.
$4.4~m/s$
(b) $130$ drops per minute.
Work Step by Step
Let the positive direction be downward with origin at the stalactite.
First, let's find the time it takes to a drop goes from the stalactite to the pool:
$x_0=0$, $x=4.0~m$, $v_0=0$, $a=g=9.81~m/s^2$
$x=x_0+v_0+\frac{1}{2}at^2$
$4.0~m=0+0t+\frac{1}{2}(9.81~m/s^2)t^2$
$8.0~m=(9.81~m/s^2)t^2$
$t=±\sqrt {\frac{8.0~m}{9.81~m/s^2}}$. The valid solution is:
$t=0.903~s$
(a) According to the given information, the second drop will detach $\frac{0.903~s}{2}=0.4515~s$ after the first drop has detached from the stalactite. That is, 1 drop for each $0.4515~s$.
$x_0=0$, $v_0=0$, $a=g=9.81~m/s^2$, $t=0.4515~s$
$x=x_0+v_0+\frac{1}{2}at^2$
$x=0+0t+\frac{1}{2}(9.81~m/s^2)(0.4515~s)^2$
$x=1.0~m$.
That is, $1.0~m$ below the stalactite or $3.0~m$ above the poll.
$v=v_0+at$
$v=0+(9.81~m/s^2)(0.4515~s)=4.4~m/s$
(b) $1~drop=0.4515~s$
$1~drop=(0.4515~s)(\frac{1~min}{60~s})$
$(1~drop)(\frac{60~s}{0.4515~s})=1~min$
$130~drops=1~min$
