In the previous problem, if $y ( 0 ) = R$ , the escape velocity is
A) $v _ { 0 } = 2 k / ( m R )$
B) $v _ { 0 } = 2 R / ( m k )$
C) $v _ { 0 } = 2 m / ( k R )$
D) $v _ { 0 } = \sqrt { 2 m / ( k R ) }$
E) none of the above

Question

Quizplus · Accepted Answer

The escape velocity is typically derived from the formula $v_0 = \sqrt{2GM/R}$, where $G$ is the gravitational constant and $M$ is the mass of the celestial body. None of the given options match this form.

In the Previous Problem, If y(0)=Ry ( 0 ) = Ry(0)=R

In the Previous Problem, If $y ( 0 ) = R$