Answer
(a) $24~m$
(b) $22~m/s$
(c) $-310~m/s^2$
Work Step by Step
Let the positive direction be downward with origin at the top of the tower.
(a) $x_0=0$, $t=2.2~s$, $v_0=0$, $a=g=9.81~m/s^2$
$x=x_0+v_0t+\frac{1}{2}at^2$
$x=0+0(2.2~s)+\frac{1}{2}(9.81~m/s^2)(2.2~s)^2$
$x=24~m$
(b) $v=v_0+at$
$v=0+(9.81~m/s)(2.2~s)=21.582~m/s\approx22~m/s$
(c) Now, the initial velocity is the final velocity of the previous step:
$\Delta x=0.75~m$, $v_0=21.582~m/s$, $v=0$, $a=g=9.81~m/s^2$
$v^2=v_0^2+2a\Delta x$
$0^2=(21.582~m/s)^2+2a(0.75~m)$
$a(-1.5~m)=(21.582~m/s)^2$
$a=\frac{(21.582~m/s)^2}{-1.5~m}=-310~m/s^2$
