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

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr