Answer
The fragment lands a distance of $~~53.1~m~~$ away from the gun.
Work Step by Step
We can find the horizontal component of velocity just before the shell explodes:
$v_x = v_0~cos~\theta_0$
$v_x = (20~m/s)~cos~60^{\circ}$
$v_x = 10~m/s$
Let the original mass of the shell be $2M$
We can find the horizontal component of the initial momentum:
$p_{xi} = 2M~v_x$
$p_{xi} = 2M~(10~m/s)$
We can find the horizontal component of velocity of the shell fragment that continues moving horizontally after the explosion:
$p_{xf} = p_{xi}$
$M~v_{xf}+M(0) = 2M~(10~m/s)$
$v_{xf} = 20~m/s$
We can find the vertical component of the initial velocity:
$v_{0y} = (20~m/s)~sin~60^{\circ}$
$v_{0y} = 17.32~m/s$
We can find the time it takes the shell to reach maximum height:
$v = v_{0y}+at$
$t = \frac{v - v_{0y}}{a}$
$t = \frac{0 -17.32~m/s}{-9.8~m/s^2}$
$t = 1.77~s$
In this time, we can find the horizontal distance that the shell travels:
$x_1 = (10~m/s)(1.77~s) = 17.7~m$
The two fragments will also take a time of $1.77~s$ to fall back down to the ground. In this time, we can find the horizontal distance that the shell fragment that continues horizontally travels:
$x_2 = (20~m/s)(1.77~s) = 35.4~m$
We can find the total horizontal distance that this fragment travels:
$d = 17.7~m+35.4~m = 53.1~m$
This fragment lands a distance of $~~53.1~m~~$ away from the gun.
