Arun
Last Activity: 6 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
Arun (askIITians forum expert)
