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)=(2hg)/(1+h/R)`
where `R` is the radius of the earth and `g` is acceleration due to gravity at earth's suface. Deduce an expression for maximum height reachhed by a rocket fired with speed `0.9` times the escape velocity.

Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from earth's surface is:

What will be the acceleration due to gravity at height h if h gt gt R . Where R is radius of earth and g is acceleration due to gravity on the surface of earth

A particle is projected vertically upwards with a velocity sqrt(gR) , where R denotes the radius of the earth and g the acceleration due to gravity on the surface of the earth. Then the maximum height ascended by the particle is

What will be the acceleration due to gravity at height h lt lt R . Where R is radius of earth and g is acceleration to gravity on the surface earth

A particle is projected vertically upwards with speed v_(0) . The drag force acting on it given by f_("drag")=mlambdav^(2) . The time when it is at maximum height is :

Given that v is speed, r is the radius and g is the acceleration due to gravity. Which of the following is dimensionless

A particle of mass m is fired vertically upward with speed v_(0) (v_(0) lt " escape speed ") . Prove that (i) Maximum height attained by the particle is H = (v_(0)^(2)R)/(2gR - v_(0)^(2)) , where g is the acceleration due to gravity at the Earth's surface and R is the Earth's radius . (ii) When the particel is at height h , the increase in gravitational potential energy is (mgh)/(1+h/R)

A body is projected vertically upwards at time "t=0" and it is seen at a height H at instants t_(1) and t_(2)seconds during its flight.The maximum height attained is (g is acceleration due to gravity

If V, R and g denote respectively the escape velocity from the surface of the earth radius of the earth, and acceleration due to gravity, then the correct equation is