A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is + 15 m at t = 2 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.

A man is standing on top of a building 100 m high. He throws two balls

1.	A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is + 15 m at t = 2 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
Answer» Solution :Let the speeds of the two balls `v _(1) and v _(2)` respectively. If `v _(2) = v, ` then `v _(1) = 2v` If `y _(1) and y _(2)` are the DISTANCE covered by the balls respectively before coming to rest, then `y_(1) = (v _(1) ^(2))/( 2G) = (4v ^(2))/( 2g) and y _(2) = (v _(2) ^(2))/( 2g) = (v ^(2))/( 2g)` `y _(1) - y _(2) = 15 m` `(4v ^(2))/(2g) - (v ^(2))/(2g) =15m` `(3v ^(2))/(2g) =15m` `therefore v ^(2) = sqrt (5m xx (2 xx 10 )) m//s ^(2)` `therefore v = 10 m//s` `therefore v _(2) = 10 m //s and v _(1) = 20 m //s` Now, `y _(1) = (v _(1) ^(2))/( 2g ) = ((20 m ) ^(2))/( 2 xx 10m15 ) = 20 m` `y_(2) = y _(1) - 15 m = 5m` If `t _(2)` is the time taken by the ball 2 to cover a distance of 5 m, then From `y _(2) = v _(2) t - (1)/(2) g t _(2) ^(2) ` `5 = 10 t _(2) - 5t _(2) ^(2) or t _(2) ^(2) + 1=0` `therefore t _(2) =1 s` Now, `t _(1)` (time taken by ball 1 to cover distance of 20 m) is 2s, time interval between the two throws `=t_(1) -t _(2) =2s -1S =1s`