A rubber ball is thrown downwards from the top of a tower with a velocity of14 m/s. A second ball is dropped from the same place 1 s later. The firstball reaches the ground in 2 s and rebounds with the same magnitude of velocity . How much later wouldthe two balls collide with each other ?

A rubber ball is thrown downwards from the top of a tower with a veloc

1.	A rubber ball is thrown downwards from the top of a tower with a velocity of14 m/s. A second ball is dropped from the same place 1 s later. The firstball reaches the ground in 2 s and rebounds with the same magnitude of velocity . How much later wouldthe two balls collide with each other ?
Answer» Solution :HEIGHT of thetower, h=Distance travelley bythe FIRST ball in 2 s(t=0 to t=2s ) `=14xx2+1/2xx9.8xx2^2=47.6 m` The second ball startsat t=1 s. From t=1 s tot=2 s, this ball comes down through a height, `h. =1/2xx9.8xx(2-1)^2=4.9m` `therefore`Att=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 thesame instant , i.e ,at t=2s is `v_(1)^(.) =-33.6 m//s.` For the second ball , att=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 totravel the height of H =42.7m is , `t=H/V=(42.7)/(43.4)=0.98 s` Therefore , theballs will collide at `T=2+t=2.98s.`