Ball-1 is dropped from the top of a building from rest. At the same moment, ball-2 is thrown upward towards ball-1 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 passed ball-2. (Assuming g=10 m/s²)

For downward motion of ball-1 from second equation of the motion,

h = ut + 12gt² ...(i)

h = 0×t + 12×10t² [∵u=0]

h = 5t² ...(ii)

For the upward motion of ball-2, from the second equation of the motion,

21 - h = vt − 12gt² ...(iii)

Given, speed of ball-2, v=14 m/s, acceleration due to gravity, g=10m/s²

Substituting these values in Eq. (iii), we get

21−h = 14t −12×10t²

⇒21−h = 14t−5t² ...(iv)

Adding Eq. (ii) and (iv), we get

21 = 14t

⇒t = 32s = 1.5s

∴ From Eq. (ii), we get

h = 5×(1.5)² = 11.25m

=45/4m

Therefore, the ball-1 will have dropped 45/4 m when it passes ball-2.