1.

A particle is given a velocity(v_(e ))/(sqrt(3)) in a vertically upward direction from the surface of the earth, where v_(e ) is the escape velocity from the surface of the earth. Let the radius of the earth be R. The height of the particle above the surface of the earth at the instant it comes to rest is :

Answer»

3R
2R
`(3R)/(4)`
`(R )/(2)`

Solution :We already KNOW that the escape VELOCITY from the SURFACE of the EARTH, `v_(e)=sqrt((2GM)/(R))`
Conserving energy: `U_(i)+K_(i)=U_(f)+K_(f)`
`-(GMm)/(R)+(1)/(2)m((v_(e))/(sqrt(3)))^(2)=-(GMm)/((R+h))`
Solving, we get `h=(R )/(2)`


Discussion

No Comment Found

Related InterviewSolutions