Ball-1 is dropped from the top of a building from rest. At the same moment, ball- 2 is thrown upward towards ball-l with a speed 14 ~m / s from a point 21 ~m below the top of building. How far will the ball-1 have dropped when it passes ball-2. (Assume, acceleration due to gravity, g=10 ~m / s ^2 .)

Step-by-step solution

Suppose that both the balls meet at a distance h from ball-l after time t from the start as shown in the figure, For downward motion of ball-1, from second equation of the motion, aligned & h=u t+ 1 2 g t^2 & h=0 t+ 1 2 10 t^2 (i) & h=5 t^2 [ u=0] aligned For upward motion of ball-2, from second equation of the motion, h=v t- 1 2 g t^2 Given, speed of ball-2,v =14 ~m / s , acceleration due to gravity, g=10 ~m / s ^2 Substituting these values in Eq. (iii), we get aligned & 21-h=14 t- 1 2 10 t^2 & 21-h=14 t-5 t^2 ali