Question

From the top of a tower 400m high a ball is dropped and at the same time another ball is thrown vertically up from the ground at a speed of 100m/s. Assuming both balls move along same line find when and where the balls meet? 
 
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Answer 1

Letting the ground to be the origin for this 1-D kinematics problem, taking upward direction as positive and downward direction as negative and writing displacement for both the balls (assuming g=10 m/(s^2) , t=0 at time of release):-
Ball1 :- x1=400 - 5*(t^2) 
Ball2 :- x2=100t - 5*(t^2) 
When both will meet x1=x2 :-
==> t = 4 seconds

Alternatively:- 
Using concepts of relative motion, relative speed of approach = 100m/s , relative acceleration = 0 , relative separation = 400m 
==> 400 = 100*t ==> t = 4 seconds
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