A man is standing on top of a building 100 m high. He throws two ball 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. At `t=2`, both the balls reach to their and second ball is `+15m`.
Q. The speed of first ball is

Explanation

Let the speed of the two balls (1 and 2) be `v_(1)` and `v_(2)` where
if `v_(1) = 2v, v_(2) = v`
if `y_(1)` and `y_(2)` and th distance covered by
the balls 1 and 2, 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)`
Since, `y_(1) - y_(2) = 15 m(4v^(2))/(2g) - (v^(2))/(2g) = 15 m` or `(3v^(2))/(2g) = 15m`
or `v^(2) = sqrt(5mxx(2xx10)) m//s^(2)`
or `v = 10 m//s`
Clearly, `v_(1) = 20m//s` and `v_(2) = 10m//s`
as `y_(1) = (v_(1)^(2))/(2g) ((20m)^(2))/(2 xx 10m15) = 20 m`
`y_(2) = y_(1) - 15m = 5m`
If `t_(2)` is the time taken by the ball 2 tonar
a distance of `5m`, then form `y_(2) = v_(2)t - (1)/(2) "gt"_(2)^(2)`
`5 = 10t_(2) - 5t_(2)^(2)` or `t_(2)^(2) - 2t_(2) + 1 = 0`
where, `t_(2) = 15`
Since `t_(1)` (time taken by ball 1 to cover distance of 20m) is 2s, time interval between the two thvws
`= t_(1) - t_(2) = 2s - 1s = 1s`