Question

What is the maximum height attained by a body projected with velocity equal to one third of the escape velocity?

Answer 1

Answer:

because with 3/4 of escape velocity the object cannot reach the height R how can it go above R

(1) 97R97R

Explanation:

According to the law of conservation of energy,

(T.E)atsurface=(T.E)atheight(T.E)atsurface=(T.E)atheight

(K.E+P.E)surface=(K.E+P.E)maxheight(K.E+P.E)surface=(K.E+P.E)maxheight

12mv2+(−GMmR)12mv2+(−GMmR) =0+(−GMmR+h)0+(−GMmR+h)

GivenV=34ve=342GMR−−−−√GivenV=34ve=342GMR

=12m×916×(2GMR)+(−GMmR)=12m×916×(2GMR)+(−GMmR)

=(−GMmR+h)=(−GMmR+h)

9GMm16R−GMmR=−GMmR+h9GMm16R−GMmR=−GMmR+h

−7GMm16R=−GMmR+h−7GMm16R=−GMmR+h

716R=1R+h716R=1R+h

⇒7R+7h=16R⇒7R+7h=16R

h=97R