Gravitation Question 117
Question: A projectile is projected with velocity $ [kv_{e}] $ in vertically upward direction from the ground into the space. ($ [v_{e}] $ is escape velocity and $ [k<1)] $ . If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be : (R = radius of earth) [Roorkee 1999; RPET 1999]
Options:
A) $ [\frac{R}{k^{2}+1}] $
B) $ [\frac{R}{k^{2}-1}] $
C) $ [\frac{R}{1-k^{2}}] $
D) $ [\frac{R}{k+1}] $
Show Answer
Answer:
Correct Answer: C
Solution:
Kinetic energy = ½mv²
$ [\frac{1}{2}m,{{(kv_{e})}^{2}}=\frac{mgh}{1+\frac{h}{R}}] $
Þ $ [\frac{1}{2}mv^{2}mgR=\frac{mgh}{1+\frac{h}{R}}] $
$ [\Rightarrow ,h=\frac{Rk^{2}}{1-k^{2}}] $
Height of Projectile from the earth’s surface = h Height from the centre of the earth
$ [r=R+h=R+\frac{Rk^{2}}{1-k^{2}}] $ By solving $ [r=\frac{R}{1-k^{2}}] $
BETA
