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

`(3H)/(4)`

Explanation

If `t` is the regular 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)`.