A body is thrown vertically upwards with a velocity v from the surface of the earth. Show that the height h up to which the body will rise is given by `h=(v^2R)/(2gR-v^2)` here ,R=radius of the earth , g=acceleration due to gravity.

An object is projected vertically upwards with a velocity u from the surface of the earth. Show that the maximum height reached by the object is h=(u^2R)/(2gR-u^2) where R is the radius of the earth. Calculate escape velocity from it.

A ball is thrown vertically upward with a speed v from a height h above the ground . The time taken for the ball to hit the ground is

A body thrown vertically upwards from a point with a velocity v_(0) rises to a maximum height and then comes back to the point. Then

Which of the following quantities is dimensionless if v = speed , r = radius of a circular path and g = acceleration due to gravity?

A body is projected vertically upwards from the surface of earth with a velocity equal to half of escape speed. What is the maximum height attained by the body ?

The mass of a planet is 4 times the mass of the earth and its radius is 2 times the radius of the earth. The acceleration due to gravity on that planet is

A rocket starts vertically upward with speed v_0 . Show that its speed v at height h is given by v_0^2-v^2=(2gh)/(1+h/R) where R is the radius of the earth.

At what height above the surface of the earth will the acceleration due to gravity be one-fourth of its value on the surface of the earth ?[Radius of the earth =R]

A lift is moving up with a constant acceleration a. A man standing on the lift, throws a ball vertically upwards with a velocity v, which returns to the thrower after a time t. show that v=(a+g) t/2 where g is the acceleration due to gravity.