An object is propelled vertically to a maximum height of $4 R$ from the surface of a planet of radius $R$ and mass $M$. The speed of the object when it returns to the surface of the planet is

Three solid spheres each of mass $1 \ kg$ and radius $2 \ m$ are arranged at the three corners of an equilateral triangle of side $10 \ m$,such that the centers of the spheres coincide with the corners of the triangle. When they are released from that position,the speed of any one sphere at the time of collision would be ($G$ is the universal gravitational constant).

$A$ rocket is fired vertically with a speed of $5 \; km/s$ from the Earth's surface. How far from the Earth's surface does the rocket go before returning? (Mass of the Earth $M_e = 6.0 \times 10^{24} \; kg$,mean radius of the Earth $R_e = 6.4 \times 10^{6} \; m$,$G = 6.67 \times 10^{-11} \; N m^2 kg^{-2}$)

$A$ body of mass $m$ falls from a height $R$ above the surface of the earth,where $R$ is the radius of the earth. What is the velocity attained by the body on reaching the ground? (Acceleration due to gravity on the surface of the earth is $g$)

$A$ body is released from a height equal to the radius $(R)$ of the earth. The velocity of the body when it strikes the surface of the earth will be: (Given $g$ = acceleration due to gravity on the earth.)

$A$ projectile is thrown straight upward from the Earth's surface with an initial speed $v = \alpha v_E$,where $\alpha$ is a constant and $v_E$ is the escape speed. The projectile travels up to a height $h = 800 \ km$ from the Earth's surface before it comes to rest. The value of the constant $\alpha$ is (Radius of the Earth $R = 6400 \ km$):