Two students are on a balcony 18.5 m above the street. one student throws a ball ver- tically downward at 13.8 m/s; at the same instant, the other student throws a ball verti- cally upward at the same speed. the second ball just misses the balcony on the way down. how far apart are the balls 0.361 s after they are thrown? the acceleration of gravity is 9.8 m/s2 .

11.24m

Explanation:

Given that:

Using the equations of motion:

To find the final velocity after the given time 0.361 second.

................ (1)

where:

Putting the values of the respective balls

Similarly  

 ∵the body is thrown against the gravity will have decelerating effect.

∵the body is still moving up with this velocity

Now, using the eq.

.............................(2)

where, the symbols have usual meaning as above and:

Putting the respective values in eq. (2)

For the body in the first case:

For the body in the second case:

∴The bodies are apart from each other from the point from which they are thrown.

i think the answer is b

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.