Ball a is dropped from the top of a building of height h at the same instant that ball b is thrown vertically upward from the ground. when the balls collide, they are moving in opposite directions, and the speed of a is twice the speed of b. at what height does the collision occur?

(2/3) times height collision occur

Explanation:

for ball A

from the kinematic equation

the distance of ball A is

= velocity ( time ) = distance

since ball is at height, the above equation changes as

for ball B

the condition of collision is

xA = xB

(given)

from the kinematic equation

the speed of the ball A is

since initial speed of the ball A is zero

, so

the speed of the ball B is

since

since

so, (2/3) times height collision occur

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.