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A body is projected vertically with speed 10 and other at same time with same speed in downward direction from top of a tower.the magnitude of the accelaration of 1st body w.r.t 2nd body is what??

A body is projected vertically with speed 10 and other at same time with same speed in downward direction from top of a tower.the magnitude of the accelaration of 1st body w.r.t 2nd body is what??

Grade:12th pass

2 Answers

Kuldeep Pal
53 Points
5 years ago
Zero:
 
As the motion is only in 1-Dimension so,There is no need for vecor sign. WE can assume as +se as vertical direction And -ve  as downwars Direction.
a(relative)= a1-a2
according to our convention vertical motion is +ve and downward motion is -ve.
So. a1=+g
       a2=-g
Hence ,
            a(relative) = a1-a2
                                     =(+g)-(-g)
                                      =g-g
                                      =0.  
                     
Raman Mishra
67 Points
5 years ago
Dear Taniya, no need to fret. This is a very simple question used for confusing the students. Since the acceleration of both the bodies will be g downwards, therefore their relative acceleration will be zero.
Taking upward direction to be positive and ground as refrence, we have:
  Acceleration of 1st body = a1 = -g
  Acceleration of 2nd body = a2 = -g
 \therefore Acceleration of 1st body wrt 2nd is given by:
              a12 = a1 – a2 = -g – (-g) = 0

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