http://oyc.yale.edu/sites/default/files/problem_set_1_solutions_1.pdf, no 5. The ball A is dropped from the building and ball B is thrown up from the ground. When they collide, the height will be h.
Then, I think, yB=h, yA=H-h. I can't understand why yA=yB=h.. Can anybody explain this to me?

Question

http://oyc.yale.edu/sites/default/files/problem_set_1_solutions_1.pdf, no 5. The ball A is dropped from the building and ball B is thrown up from the ground. When they collide, the height will be h.
Then, I think, yB=h, yA=H-h. I can't understand why yA=yB=h.. Can anybody explain this to me?

anonymous · Answer

dear friend it is expicitly based upon the concept of relative velocity..
relative to A ball B would exactly be travelling at the pace Vb -Va.?so time required for collision in the frame of A would be H/(Va-Vb)..once the time is know the problem becomes a kid's stuff,,apply second equation of motion
s=ut+o.5gt^2 for ball B?

theraggedydoctor · Answer

;)))