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A man of mass 60 kg is standing in a gravity free zone at a height of 12 m above the floor . He throws a stone of mass 0.5 kg downwards with a speed of 4 m/s . When the stone reaches the floor , the distance of the man above the floor will be?

A man of mass 60 kg is standing in a gravity free zone at a height of 12 m above the floor . He throws a stone of mass 0.5 kg downwards with a speed of 4 m/s . When the stone reaches the floor , the distance of the man above the floor will be?

Grade:10

1 Answers

KSK
17 Points
9 years ago
H=12
M1=60KG, V1=?,M2=0.5KG, V2= 4 m/s
 
USING CONSERVATION OF MOMENTUM
INITIAL MOMENTUM= FINAL MOMENTUM
0 = M1*V1+M2*V2
M1*V1=(-M2*V2)
60*V1=(-0.5*4)
V1= (-1)/15 = 1/15  m/s (IN UPWARD DIRECTION)
 
(T) TIME TAKEN BY STONE TO REACH GROUND = DISTANCE/SPEED
  T= 12/4 = 3 sec
(D) DISTANCE COVERED IN UPWARD DIRECTION BY MAN WHEN STONE HITS GROUND=
3 * (1/15)
D= 1/5= 0.2m 
DISTANCE BETWEEN GROUND AND MAN = 12+0.2 = 12.2 m

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