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))`

Options

A. `10m`
B. `20m`
C. `15m`
D. `5m`

Answer

B

Explanation

`u=20ms^(-1)`
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`