Question

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 ?

Answer 1

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

Answer 2

`y_(cm) = 7.05 m`

Step-by-step explanation:

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

`y_(cm) = (m_1y_1 + m_2y_2)/(m_1 + m_2)`

here we know that

`y_1 = y - (1)/(2)gt^2`

`y_1 = 25 - (1)/(2)(9.81)(2^2)`

`y_1 = 5.38 m`

Similarly for another ball we have

`y_2 = v_y t - (1)/(2)gt^2`

`y_2 = 15(2) - (1)/(2)(9.81)(2^2)`

`y_2 = 10.38 m`

now for COM for above two masses

`y_(cm) = (0.50(5.38) + 0.25(10.38))/(0.50 + 0.25)`

`y_(cm) = 7.05 m`