At the same instant that a 0.50-kg ball is dropped from 25 m above earth, a second ball, with a mass of 0.25 kg, is thrown straight upward from earth's surface with an initial speed of 15 m/s. they move along nearby lines and pass each other without colliding. at the end of 2.0 s, the height above earth's surface of the center of mass of the two-ball system is:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

Motion of 0.50 kg ball:

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

                 s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

Motion of 0.25 kg ball:

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

The center of mass of the two-ball system is 7.05 m above ground.  

they should be similar or the same

firearm identification is the process of analyzing the bullets and cartridge cases left at a crime scene to determine if they came from a particular firearm. marks on the bullets and cartridges may be common to every firearm of that type. these are called class characteristics.