Question

At the same instant that a 0.50-kg ball is dropped from 25m above Earth,? At the same instant that a 0.50-kg ball is dropped from 25m aboveEarth, a second ball, with a mass of 0.25 kg, is thrown straight upward from Earth’ssurface with an initial speed of 15m/s. They move along nearby lines and pass each other without colliding.At the end of 2.0 s the velocity of the center of mass of the two-ball system is:

Answer 1

I hope it is clearly visible.. Velocity of the center of mass of 2-ball system is - 11.54m/s. Minus indicates, velocity direction is in downward direction.

Answer 2

Following are the calculation to the given question:

Ball 1's Y-coordinate:

$\to y_{1f} = y_1 + v_1 t -\frac{1}{2} gt^2$

           $= 25+0- \frac{1}{2}\times 9.8 \ \frac{m}{s^2} \times (2 s)^2\\\\= 25- \frac{1}{2} \times 9.8 \ \frac{m}{s^2} \times 4\ s^2\\\\= 25- 9.8\ m \times 2\\\\= 25- 19.6\ m \\\\= 5.4\ m$

As a result, the distance inside the ball's y-coordinate is 5.4 m.

Ball 2's Y-coordinate:

$\to y_{2f} = v_2 t -\frac{1}{2} gt^2$

          $=15 \times 2- \frac{1}{2}\times 9.8 \ \frac{m}{s^2} \times (2 s)^2\\\\=30- \frac{1}{2}\times 9.8 \ \frac{m}{s^2} \times 4 s^2\\\\=30- 9.8 \ m \times 2 \\\\=30- 19.6 \ m\\\\= 10.4\ m$

As a result, the distance in ball 12's y-coordinate is 10.4 m.

approximate distance from the center of mass:

$\to y= \frac{m_1y_1+ m_2 y_2}{ m_1+m_2}$

       $=\frac{(0.5\ kg \times 5.4\ m) +(0.25\ kg \times 10.4\ m)}{(0.5+0.25) \ kg}\\\\ =\frac{(2.7 \ kg- m) +(2.6\ kg- m)}{(0.5+0.25) \ kg}\\\\ =\frac{5.3\ kg- m}{(0.75 ) \ kg}\\\\=7.066\ m \approx 7.1 \ m$

As a result, the height above the earth's surface from the center of the ball is 7.1 m.

Learn more:

brainly.com/question/14298716