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 ?

A stone is fallen freely from a tower and after n seconds another ston

1.	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 ?
Answer» Solution :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)`