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 ?

In your question when a 0.5kg ball is dropped from 25m above Earth, a second ball with a mass of 0.25 kg is thrown straight upward from the earth with an initial speed of 15m/s. I think the center of mass is 6.25 kg m/s

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aristocles aristocles · Answer 2 · 2019-12-02T03:34:53+01:00

Answer:

[tex]y_{cm} = 7.05 m[/tex]

Explanation:

As we know that the position of center of mass of the system of two masses is given as

[tex]y_{cm} = \frac{m_1y_1 + m_2y_2}{m_1 + m_2}[/tex]

here we know that

[tex]y_1 = y - \frac{1}{2}gt^2[/tex]

[tex]y_1 = 25 - \frac{1}{2}(9.81)(2^2)[/tex]

[tex]y_1 = 5.38 m[/tex]

Similarly for another ball we have

[tex]y_2 = v_y t - \frac{1}{2}gt^2[/tex]

[tex]y_2 = 15(2) - \frac{1}{2}(9.81)(2^2)[/tex]

[tex]y_2 = 10.38 m[/tex]

now for COM for above two masses

[tex]y_{cm} = \frac{0.50(5.38) + 0.25(10.38)}{0.50 + 0.25}[/tex]

[tex]y_{cm} = 7.05 m[/tex]