Answer
(a) $9.81~m/s^2$ and points downward.
(b) $13.9~m$
(c) $2.21~s$
(d) $16.5~m/s$
Work Step by Step
(a) After being dropped, the shell is only subjected to the gravity (free fall). So the magnitude of the acceleration is $9.81~m/s^2$ and points downward.
(b) Let the positive direction be upward with origin at the ground.
At the maximum height $v=0$
$v_0=5.20~m/s$, $v=0$, $x_0=12.5~m$, $a=-g=-9.81~m/s^2$
$v^2=v_0^2+2a\Delta x$
$0^2=(5.20~m/s)^2+2(-9.81~m/s^2)\Delta x$
$(19.62~m/s^2)\Delta x=(5.20~m/s)^2$
$\Delta x=\frac{(5.20~m/s)^2}{19.62~m/s^2}=1.38~m$
$\Delta x=x-x_0$
$x=\Delta x+x_0=1.38~m+12.5~m=13.9~m$
(c) When the shell reaches the ground $x=0$.
$x_0=12.5~m$, $x=0$, $v_0=5.20~m/s$, $a=-g=-9.81~m/s^2$
$x=x_0+v_0t+\frac{1}{2}at^2$
$0=12.5~m+(5.20~m/s)t+\frac{1}{2}(-9.81~m/s^2)t^2$
$(4.905~m/s^2)t^2+(-5.20~m/s)t+(-12.5~m)=0$
This is a quadratic equation for $t$.
$t=\frac{-(-5.20~m/s)±\sqrt {(-5.20~m/s)^2-4(4.905~m/s^2)(-12.5~m)}}{2(4.905~m/s^2)}$
$t_1=2.21~s$
$t_2=-1.15~s$ (not valid).
It takes $2.21~s$ for the shell to reach the ground.
(d) $t=2.21~s$, $v_0=5.20~m/s$, $a=-g=-9.81~m/s^2$
$v=v_0+at$
$v=5.20~m/s+(-9.81~m/s^2)(2.21~s)$
$v=-16.5~m/s$
$speed=|v|=16.5~m/s$
