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debmalya.roychoudhuri (96)

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1.a ballooon starts rising from the ground with an acceleartion of 1.25m/sec^2 .after 8secs a stone is droopped from the balloon the stone will :


a cover a distance of 40 m   b.have a displacement of 50 m


c.reach the ground in 4 secs  d. begin to move down after being released


which is/are the coorrect answers plzzz explin


2.an aerplane flying at a constant velocity releases a bomb.as the bomb drops awayfrom the aeroplane ,


a.it will always be vertically below the pl;ane   


b.it will aways be vertically below the plane only if it was flying horizonataallly 


it will gradually fall behind the plane if it was flying horizonatlly


plzzz answer with explanation ill be grateful and rate ur explanations


thank u (answers awaited)


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varun.tinkle (1290)

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1 ST PART

SINCE IT IS ACC UPWARDS WITH AN ACC OF 1.25

THE VELOCITY AT THE END OF 8 S

=8*1.25 =10 M/S

SO WHEN THE STONE IS RELEASED ITS VELOCITY IS 10 MS AND ACC -(10-1.25) (ASSUMING DOWNWARDS AS NEGATIVE AXIS)     ( AND ALSO 1.25  SOINCE IF ULEAVE THE OBJECT IN A NON GRAVITY FRAME IT WILL ACCLERATE UPWARDS BUT NOW THE RESULTANT ACC WILL BE TAKEN INTO ACCOUNT)

AND THE DISTANCE OF THE STONE FROM THE GROUND IS

1/2AT^2=40m

THE OBJECT WILL NOT MOVE IMMEDIATELY DOWNWARDS SINCE IT HAS A VERTICAL VELOCITY IN THE UPWARDS DIRECTION


AND THE REST U CAN FIND OUT ON UR OWN THE TIME TAKEN WILL BE

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varun.tinkle (1290)

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2nD PART
AT THIS POINT UNDERSTAND THAT THE HORIZONTAL COMPONENT OF VELOCITY IS INDEPENDENT OF VERTICAL COMPONET SO AND IF THE STONE IS KEPT IN A SYSTEM MOVING WITH A VELCOITY ITS INITIAL VELOCITY WILL BE IN THAT DIRECTION ONLY
APPLYING THIS PRINCIPLE
WE GET THAT
IT WILL BE b
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