Question

two particle A and B are projected in air.B is thrown vertically and Ais thrown vertically up.what is the separation between them after 1 sec​

Answer 1

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Explanation:Solution :

Method -1: Position of A aftre 1 s

SaA=ut−12gt2

=5t−12×10×t2=5×1−5×12=5−5=0

i.e., the particle will return to ground at t=1s

SB=ut−12gt2=10×1−12×10×12=10−5=5m

Hence, separation between A and B, SB−SA=5m

Method-2: Relative velocity method:

a→BA=v→A=10−5=5ms−1

Hence, relative separation between particles

becuases→BA(∈1s)=v→BA×t=5×1=5m

becuase Distance between A and B after 1s=5m.