Two cannon balls weighing 13 kg and 19 kg are chained together and fired horizontally with a velocity of 165 m/s from the top of a 16-m wall. The chain breaks during the flight of the cannonballs and the 13-kg cannonball strikes the ground at t = 1.5 s, at a distance of 240 m from the foot of the wall, and 7 m to the right of the line of fire. Determine the position of the other cannonball at that instant. Neglect the resistance of the air.

Let's assume the height is the y-direction, the line of fire is in x-direction. Since there is no external force acting in z-direction, z-coordinate of the center of mass of he ball should be zero. In order to make z-coordinate zero other ball should fall symmetrically with respect to z-axis. So, z-coordinate of 19 kg ball = -7 m.

Hence, 7 m to the left of the line of fire.

The balls do not have any external force in x-direction. Thus, in x-direction, the center of mass should move with constant velocity. x-coordinate of center of mass at t = 1.5 s is:

[tex]x_{CM}= 165*1.5 = 247.5 \:\:m[/tex]

[tex]x_{CM}=\frac{m_1x_1+m_2x_2}{m_1+m_2}\\ \\247.5=\frac{13*240+19*x_2}{13+19}\\\\x_2=(247.5*29-13*240)/19=213.55 \:\:m[/tex]

Hence, 213.55 m from the foot of the wall.

Height fallen by the center of mass at t = 1.5 s.

[tex]h=\frac{1}{2} gt^2=\frac{1}{2}*10*(1.5)^2=11.25\:\:m[/tex]

Hence, y-coordinate of the center of mass is:

[tex]y_{CM}=16-11.25=4.75\:\:m\\\\y_{CM}=\frac{m_1y_1+m_2y_2}{m_1+m_2} \\\\4.75=\frac{13*0+19*y_2}{13+19}\\\\y_2=(4.75*29)/19=7.25\:\:m[/tex]

Hence, 7.25 m above the ground.