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# A stone is dropped from rest from top of a cliff  A second stone is thrown vertically down with a velocity of30m/s two seconds later. At what distance from the top of the cliff do they meet?

Arun
25763 Points
2 years ago
Dear student

The to a stone's meet at a distance S from top of Cliff t
seconds after first stone is dropped.

For 1st stone S = 1/2 gt²

For 2nd stone S = u( t -2 ) + 1/2 g(t-2)²

now equate the both distance

1/2 gt² = ut – 2u + 1/2 gt² – 2gt + 2g

0 = ( u - 2g) t - 2(u- g )

t = 2( u - g ) /u - 2g = 2(30 - 10 ) /30 - 20 = 4 sec

t = 4 second

distances S at which they meet = 1/2 * gt²

= 1/2 * 10 * 16

= 80 m from top of Cliff

Regards