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InterviewSolution
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An object is thrown vertically upwards with velocity of 20ms^(-1). At what height will its kinetic energy and potential energy be equal? (g=10ms^(-2)) |
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Answer» `10m` As per the law of CONSERVATION of MECHANICAL energy, `(("Mechanical"),("energy of the"),("OBJECT on ground"))=(("Mechanical energy"),("of the object at"),("maximum height"))` `therefore E_(k_(1))+E_(p_(1))=E_(k_(2))+E_(p_(2))` `therefore (1)/(2)m u^(2)+0=0+mgh` (`because` Potential energy of the object on ground `E_(p_(1))=0` and kinetic energy of the object at the maximum height `E_(k_(2))=0`) `therefore h=(u^(2))/(2g)=((20)^(2))/(2xx10)=20m` |
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