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

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6 cos 34, 6 sin 34 trig