Q.

From the top of a tower, a stone is thrown up and reaches the ground in time t1 . A second stone is thrown down with the same speed and reaches the ground in time t2 . A third stone is released from rest and reaches the ground in time t3 then the relation between them is

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a

t3=t1+t2

b

t3=t1t2

c

t3=t1+t22

d

t3=t1t2

answer is B.

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Detailed Solution

h – Height of the tower

u – Velocity of projection of the stone

when stone is projected  vertically upwards. Taking vertically downwards motion of stone from the top of tower to the grand we have

u=u,a=g,s=handt=t1

S=ut+12at2

h=ut1+12×g×t12(1)

when stone is projected vertically down ward motion of stone from top of tower to ground we have,

u=u,a=g,s=handt=t2

S=ut+12at2

h=ut2+12×g×t22(2)

when stone falls freely

u=0,a=g,s=h

h=0+12gt32(3)

multiplying (1) byt2 and (2) byt1 and then adding we get

h=ut1+12×g×t12×t2

t2h=ut1t2+12gt12t2

t1h=ut1t2+12gt22t1_

(t2+t1)h=12gt1t2(t1+t2)

h=12gt1t2(t1+t2)_

(t1+t2)

h=12gt1t2

substitute ‘h’ value in equation (3) then

12gt1t2=12gt32

 t1t2=t32t3:t1t2

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