An object is allowed to fall from a height $R$ above the earth,where $R$ is the radius of the earth. Its velocity when it strikes the earth's surface,ignoring air resistance,will be:

$A$ body (mass $m$) starts its motion from rest from a point distant $R_0$ $(R_0 > R)$ from the centre of the Earth. The velocity acquired by the body when it reaches the surface of the Earth will be ($G =$ universal constant of gravitation,$M =$ mass of Earth,$R =$ radius of Earth).

Two planets have masses $M$ and $16\, M$ and their radii are $a$ and $2\, a$ respectively. The separation between the centres of the planets is $10\, a.$ $A$ body of mass $m$ is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach the surface of the smaller planet,the minimum firing speed needed is

$A$ body is thrown from the surface of the earth with velocity $V \ m/s$. The maximum height above the earth's surface up to which it will reach is ($R =$ radius of earth,$g =$ acceleration due to gravity).

Suppose two planets (spherical in shape) of radii $R$ and $2R$,but mass $M$ and $9M$ respectively have a centre-to-centre separation $8R$ as shown in the figure. $A$ satellite of mass $m$ is projected from the surface of the planet of mass $M$ directly towards the centre of the second planet. The minimum speed $v$ required for the satellite to reach the surface of the second planet is $\sqrt{\frac{a}{7} \frac{GM}{R}}$,then the value of $a$ is $....$
[Given: The two planets are fixed in their position]

Given that the mass of the Earth is $M$ and its radius is $R$. $A$ body is dropped from a height equal to the radius of the Earth above the surface of the Earth. When it reaches the ground,its velocity will be: