A stone is fallen freely from a tower and after n seconds another stone is thrown upwards with u m/s, then at what distance from top of tower second stone will overtake first stone ?

Explanation

Suppose after t time of free fall of first stone, second stone will meet first stone.
`therefore` Distance covered by first stone in .t. time,
`d _(1) = (1)/(2) g t ^(2) " "…(1)`
and second stone will meet at (t-n) second to first stone.
`therefore` Distance covered by second stone in `(t-n)` seconds,
`d _(2) = u (t -n) + 1/2 g (t -n) ^(2)`
`therefore` The distance covered by first stone in t s = The distance covered by second stone in `(t-n)` s
`1/2 g t^(2) = u (t-n) + 1/2 g (t -n)^(2)`
`therefore 1/2 g t ^(2) = u (t-n) + 1/2 g (t ^(2) - 2tn + n^(2))`
`therefore 1/2 g t ^(2) = ut - un + 1/2 g t ^(2) - gtn + 1/2 gn ^(2)`
`therefore 0= ut - g tn + 1/2 gn ^(2) - un`
`therefore g t n- ut = 1/2 gn ^(2) - un`
`therefore t (gn - u) =n ((1)/(2) gn - u )`
`therefore t = (n ((1)/(2) gn -u ))/( gn - u )`
Distance h covered by first stone in t time, then ` h = 1/2 g t ^(2)`
`therefore h = 1/2 g [ (n ((1)/(2)gn - u ))/( gn - u ) ]^(2)`
From the top of the tower, at h distance first stone will overtake the second stone.
`therefore ` Distance from top of the tower,
`1/2 [ (n ((1)/(2) gn - u ))/( gn - u ) ] ^(2)`