Please correct Q- 109 A stone with a velocity of 9.8 m / Sec is projected vertically upward .At which height will be Potential energy of body half of the kinetic energy . ( a ) 3.27 meter ( b ) 2.40 meter ( c ) 2.45 meter ( d ) None Solution :- v^2=u^2-2 g h P . E . = 1/2 K . E . m g h = 1/4 mv^2 4m g h = m ( u^2-2 g h ) u^2=6 g h h = u^2/6g=〖(9.8)^2/6x9.8 =1.64 meter
Please correct Q- 109
A stone with a velocity of 9.8 m / Sec is projected vertically upward .At which height will be Potential energy of body half of the kinetic energy .
( a ) 3.27 meter ( b ) 2.40 meter ( c ) 2.45 meter ( d ) None
Solution :- v^2=u^2-2 g h
P . E . = 1/2 K . E .
m g h = 1/4 mv^2
4m g h = m ( u^2-2 g h )
u^2=6 g h
h = u^2/6g=〖(9.8)^2/6x9.8
=1.64 meter