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.

Explanation

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`