Question

A ball is dropped from the top of a building of height 80 m. at same instant another ball is thrown upwards with speed 50 m/s from the bottom of the building. the time at which balls will meet is

Answer 1

Let u1 and u2 be the initial velocities of balls 1 & 2.
h be the height at which both balls cross each other. Therefore distance travelled by ball 2 if given by:
h=50t-½gt²
ie. h=50t-4.905t²
For ball 1
Displacement is given by:
80-h=0t-½gt²
ie. 80-h=-4.905t²
Using these two equations we get 
80-50t=0

Hence, t=1.6 sec

Answer 2

Step-by-step explanation:

1.6seconds after release.

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=80 - 5*(t^2)

Ball2 :- x2=50t - 5*(t^2)

When both will meet x1=x2 :-

==> t = 1.6 seconds

Alternatively:-

Using concepts of relative motion, relative speed of approach = 50m/s , relative acceleration = 0 , relative separation = 80m

==> 80 = 50*t ==> t = 1.6 seconds