At the same instant that a 0.50 kg ball is dropped from 25 m above earth, a second ball, with a mass of .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 the 2s:

a. what is the height above earth's surface of the center of mass of the two-ball system?

b: what is the velocity of the center of mass of the two-ball system?

c: what is the magnitude of the acceleration of the center of the mass of the two-ball system?

A. 7.1m

B. 3.55m/s

C. 1.775m/s^2

Explanation:

First step is to identify given parameters;

Ball 1: m₁ = 0.5kg, u (initial velocity) =0, t = 2seconds

Ball 2: m₂ = 0.25kg, u = 15m/s, t = 2seconds

Second step: we determine the y-coordinate of ball 1 after 2 seconds, using the equation of motion under gravity as shown below;  

Recall, that the ball was thrown from a height of 25m, total y-coordinate of ball 1 after 2 seconds becomes 25m +(-19.6m)

Third step: we determine the y-coordinate of ball 2 after 2 seconds

Fourth step: we determine the y-component of the center mass of the two balls

y = 7.1m

Fifth step: we solve B part of the question; velocity of the center mass of the two balls

velocity = 3.55m/s

Sixth step: we solve C part of the question; acceleration of the center mass of the two balls

acceleration = 1.775 m/s^2

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

v=ir = 3x15 =45 v