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

Options

A. R/2
B. 16R/9
C. R/3
D. R/8

Answer

C

Explanation

`v=50/100 V_(e)=1/2 sqrt((2GM)/R)`
Applying energy conservation
`rArr -(GMm)/R+1/2 mv^(2)=-(GMm)/((R+h))`
`v^(2)=(2GM)/R-(2GM)/(R+h)rArr 1/4.(2GM)/R=2GM(1/R-1/(R+h))rArr 1/(4R)=h/(R(R+h))`
`rArr R+h=4h rArr h=R//3`