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Byju's Answer

Standard XII

Physics

Relative Velocity

A stone dropp...

Question

A stone dropped from a Minar of height h and it reaches after t seconds on Earth. From the same Minar if two stones are thrown with the same velocity u and they reach the Earth surface after t1 and t2 seconds....then:
a) t=t1-t2
B) t= (t1+t2)/2
C) t= (t1t2)^{1/2
d) t =( t₁t₂)}^{^2}

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Solution

T=(t1+t2)/2
apply v²=u²+2as. ( u will find that all three stones have same velocity on reaching the ground )
then apply v=u+at with proper sign convention
u get v=-u+gt1. (i)
v=u+gt2. (ii)
v=gt. (iii)
add 1 & 2
u get 2v=g(t1+t2)
multiply eqn 3 by two
u get 2gt=g(t1+t2)
(t1+t2)/2=t

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T=(t1+t2)/2 apply v²=u²+2as. ( u will find that all three stones have same velocity on reaching the ground ) then apply v=u+at with proper sign convention u get v=-u+gt1. (i) v=u+gt2. (ii) v=gt. (iii) add 1 & 2 u get 2v=g(t1+t2) multiply eqn 3 by two u get 2gt=g(t1+t2) (t1+t2)/2=t

T=(t1+t2)/2
apply v²=u²+2as. ( u will find that all three stones have same velocity on reaching the ground )
then apply v=u+at with proper sign convention
u get v=-u+gt1. (i)
v=u+gt2. (ii)
v=gt. (iii)
add 1 & 2
u get 2v=g(t1+t2)
multiply eqn 3 by two
u get 2gt=g(t1+t2)
(t1+t2)/2=t