A rubber ball is thrown vertically downwards from the top of a tower with an initial velocity of `14"m.s"^(-1)` . A second ball is dropped 1 s later from the same place. In 2 s the first ball reaches the ground and rebounds upwards with the same velocity. When will they collide with each other?

Explanation

Height of the tower h = distance covered by the first ball in 2s = `14 xx 2 +(1)/(2) xx 9.8 xx(2)^(2) = 47.6` m
`[because h = ut+(1)/(2)"gt"^(2)]`
Velocity of the first ball just before touching the ground is v = 14 +`9.8xx2 = 33.6 "m.s"^(-1)`
Hence its velocity just after bouncing = `33.6 "m.s"^(-1)`
Downward displacement of the second ball in 1 s,
`x = (1)/(2) xx 9.8 xx (1)^(2) = 4.9` m
Velocity of second ball after 1s = `9.8 xx 1 = 9.8 "m.s"^(-1)`
Hence distance between the two balls 2s after the projection the two first ball = 47.6 - 4.9 = 42.7 m.
Let the two balls collide with each other t s after the first ball bounces off the ground .
Upward displacement of the first ball in t s,
`x_(1) = 33.6t -(1)/(2) xx 9.8 xx t^(2) = 33.6 t -4.9t^(2)`
Downward displacement of the second ball in t s,
`x_(2) = 9.8 t + (1)/(2) xx 9.8 xx t^(2) = 9.8t +4.9t^(2)`
Now, `x_(1)+x_(2) ` = 42.7
or, `" " 33.6t-4.9t^(2)+9.8t+4.9t^(2) = 42.7 " ""or", 43.4t = 42.7`
`:. " " t = (42.7)/(43.4) = 0.98 s.`