A rubber ball is thrown downwards from the top of a tower with a velocity of 14 m/s. A second ball is dropped from the same place 1 s later. The first ball reaches the ground in 2 s and rebounds with the same magnitude of velocity . How much later would the two balls collide with each other ?

Explanation

Height of the tower,
h=Distance travelley by the first ball in 2 s (t=0 to t=2s )
`=14xx2+1/2xx9.8xx2^2=47.6 m`
The second ball starts at t=1 s. From t=1 s to t=2 s, this ball comes down through a height,
`h. =1/2xx9.8xx(2-1)^2=4.9m`
`therefore`At t=2 s, the distance between the two balls,
`H=h-h.=47.6-4.9=42.7 m `
For the first ball, velocity at t=2s is,
`v_1=14+9.8xx2=33.6 m//s`
As it rebounds with the same instant , i.e ,at t=2s is
`v_(1)^(.) =-33.6 m//s.`
For the second ball , at t=1 s, the velocity u=0.
So, at t=2s , the velocity,
`v_(2)=0+9.8xx(2-1)=9.8m//s`
`therefore` Relative velocity of the second ball with respect to the first at t=2 s is,
`V=v_(2)-v_(1)^(.)=9.8 -(-33.6)=43.4 m//s`
Relative acceleration =g-g=0. So, for relative motion , the velocity V is uniform . Thus time required to travel the height of H =42.7m is ,
`t=H/V=(42.7)/(43.4)=0.98 s`
Therefore , the balls will collide at `T=2+t=2.98s.`