The balls are released from the top of a tower of heigh H at regular interval of time. When first ball reaches at the grund, the nthe ball is to be just released and ((n+1))/(2) th ball is at some distance h from top of the tower. Find the value of h.

The balls are released from the top of a tower of heigh H at regular i

1.	The balls are released from the top of a tower of heigh H at regular interval of time. When first ball reaches at the grund, the nthe ball is to be just released and ((n+1))/(2) th ball is at some distance h from top of the tower. Find the value of h.
Answer» Solution :If `t` is theregular interval of time at which balls are thrown then `((n+1)/(2)` th ball will reach at height `KH` in time ` [(( n+1))/(2)-1]xxt`. because first ball is thrown at time ZERO. ` kH=(1)/(2) g [((n+1)/(2)-1)t]^(2)`(i) and `H=(1)/(2) g (n-1)t]^(2)` (ii) Fromequation (i) ` (2Hk)/(8) =((n-1)/(2))^(2)t^(2), (n-1)^(2)=(8kH)/(g t^(2))` From equation (ii),`(n-1)^(2)=(2H)/(g t^(2))` Comaring, `k=(1)/(4)` HEITHT `=H-(H)/(4) =(3H)/(4)`.