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A particle is projected vertically upwards with a velocity sqrt(gR), where R denotes the radius of the earth and g the acceleration due to gravity on the surface of the earth. Then the maximum height ascended by the particle is |
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Answer» `R_(e )` At maximum height the KINETIC energy becomes ZERO. Using conservation of energy, `(mV^(2))/(2)-(GM_(e)m)/(R_(e))=-(GM_(e)m)/(r)` `IMPLIES (mgR_(e))/(2)-(GM_(e)m)/(R_(e))=-(GM_(e)m)/(r)(GM_(e)=gR_(e)^(2))` `implies(mgR_(e))/(2)-mgR_(e)=-(gR_(e)^(2)m)/(r)implies(mgR_(e))/(2)=(gR_(e)^(2)m)/(r)` `implies r=2R_(e)(r=R_(e)+h) therefore` The height attained = `R_(e)` |
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