Question

At the same instant that a 9.9 kg ball is dropped from 76 m above the Earth's surface, a second ball, with a mass of 15.0 kg, is thrown straight upward from the Earth's surface with initial velocity of 8.86 They move along nearby vertical paths and pass without colliding. Calculate the position relative to the surface of the Earth of the center of mass of the two-ball system after 6.57 s. A negative answer is okay, can you interpret that physically? NOTE: Be sure to include units in your answer. VOM

Answer 1

Answer: - 146.22 m

Solution is given below.

Vertical displaceme ball displacement of 9.9kg after 7-6.57 s. is S = ut + 12gt² 49.8 x(6-57) go acceleration due to graving, 9.8 m/s2 e initial Tot 2 S, = 211.50 m the 9.9 ball comes to earth sesface after 6.575, and net mass 9.9 kg displacement from earth sonface = 76-211.50 -1355 vertical displacement of 15.0k after 6.5ts in S2 unt = M₂ - 8.8 ms = 8.66*657 - **%Ex(6.53)? Sz= –153.29 - & gt² i position of conter of mass Tan= 9.9x61355 ) + 15x6153-29) 9.9+15 Tem=-146.22m

The negative sign indicates that the position of center of mass is below the earth surface.

Thank you