A particle of mass m is projected upwards with velocity `v=(v_(e))/(2)`, where `v_(e)` is the escape velocity then at the maximum height the potential energy of the particle is : (R is radius of earth and M is mass of earth)

A particle of mass m is projected upword with velocity v=(v_(e))/(2) ( v_(e)) escape of the particle is

A particle of mass 'm' is projected with a velocity v = k V_e (k <1) from the surface of the earth.(Ve = escape velocity)The maximum height above the surface reached by the particle is:

A particle of mass m is projected upwards from the surface of the earth with a velocity equal to half the escape velocity. (R is radius of earth and M is mass of earth)Calculate the potential energy of the particle at its maximum. Write the kinetic energy of the particle when it was at half the maximum height.

The escape velocity for the earth is V_(e) . The escape velocity for a planet whose radius (1)/(4) th radius of earth and mass haif that of earth is

A ball of mass m is fired vertically upwards from the surface of the earth with velocity nv_(e) , where v_(e) is the escape velocity and nlt1 . Neglecting air resistance, to what height will the ball rise? (Take radius of the earth=R):-

A ball of mass m is fired vertically upwards from the surface of the earth with velocity nv_(e) , where v_(e) is the escape velocity and nlt1 . Neglecting air resistance, to what height will the ball rise? (Take radius of the earth=R):-

A ball of mass m is fired vertically upwards from the surface of the earth with velocity nv_(e) , where v_(e) is the escape velocity and nlt1 . Neglecting air resistance, to what height will the ball rise? (Take radius of the earth=R):-

A projectile is fired vertically upwards from the surface of the earth with a velocity Kv_(e) , where v_(e) is the escape velocity and Klt1 .If R is the radius of the earth, the maximum height to which it will rise measured from the centre of the earth will be (neglect air resistance)