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A stone of mass 10 Kg is dropped from a tower of height 122.5 m. After 1 second, if the gravitational force becomes zero. (Neglect air resistance) take g = 9.8ms^-1.
Find :
i) Velocity of the body after 2s is _____
ii) Neglecting the air resistance, the net force acting on the body is ____
iii) Distance covered in the 3rd second of its journey _____
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1. S=UT+1/2gt^2
=2u+1/2*9.8*4
122.5 =2u+19.6
122.5-19.6/2=u
U=51.45m/sec
2. Net Force acting on the body=mg
=10*9.8=98N
3. S=ut+1/2gt^2
51.45*3+1/2*9.8*9= 198.45
=2u+1/2*9.8*4
122.5 =2u+19.6
122.5-19.6/2=u
U=51.45m/sec
2. Net Force acting on the body=mg
=10*9.8=98N
3. S=ut+1/2gt^2
51.45*3+1/2*9.8*9= 198.45
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