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CBSE Class 11-science Answered

A stone A is dropped from the top of a tower 20 m high. simultaneously another stone B is thrown up from the bottom of this tower so that it can can reach just the top of the tower what is the distances of the stone from ground while they pass each other.
Asked by tps.mjmdr | 09 Jul, 2018, 05:28: PM
answered-by-expert Expert Answer
projection velocity u of ball B from bottom is calculated using the formula " v2 = u2 - 2gh " , with final velocity v =0
 
begin mathsize 12px style u space equals space square root of 2 cross times g cross times h end root space equals space square root of 2 cross times 10 cross times 20 end root space equals space 20 space m divided by s end style
( for conveneience of calculation, I assume g = 10 m/s2 )
 
Let t be the time for both the balls to meet above ground.
 
Ball-A is dropped from 20 m height, hence vertical distance hA travelled by ball is given by
 
begin mathsize 12px style h subscript A space equals space 1 half space g space t squared space equals space 5 space t squared end style ...............(1)
Vertical distance hB travelled by ball-B is given by,
 
begin mathsize 12px style h subscript B space equals space u cross times t space minus space 1 half g cross times t squared space equals space 20 cross times t space minus space 5 cross times t squared space end style ............(2)
begin mathsize 12px style h subscript A space plus space h subscript B space equals space 20 cross times t space space equals space 20 space space end style
hence t = 1 s ; 
 
hB is the distance required to be calculated. From eqn.(2) we get hB = 15 m
Answered by Thiyagarajan K | 11 Jul, 2018, 03:32: PM
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