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Grade: 12th pass
        
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?
5 months ago

Answers : (1)

Arun
20876 Points
							
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)
5 months ago
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