Question

A body is projected vertically upwards from the surface of the earth with a velocity equal to half the escape velocity. If \(R\) is the radius of the earth, maximum height attained by the body from the surface of the earth is

• A. R/6
• B. R/3
• C. 2R/3
• D. R

Answer 1

The Correct Option is B

Applying conservation of mechanical energy,
$\frac{1}{2} m\left(\frac{v_{e}}{2}\right)^{2}=\frac{m g h}{1+\frac{h}{R}} $
or $ \frac{v_{e}^{2}}{8}=\frac{g h}{1+\frac{h}{R}} $
?? $ \frac{2 g R}{8}=\frac{g h}{1+h / R} \ldots$ (i)
$\left(\because v_{e}=\sqrt{2 g R}\right)$
On solving E (i), we get $h=\frac{R}{3}$.

Answer 2

The correct option is (B)

The maximum height obtained by a projectile
\(h = \frac{v^{2}R}{2gR−v^{2}​}\)     …..(i)
The velocity of the body = half the escape velocity
\(v=\frac{v_{e}}{2}​​\)
Or \(v=\frac{\bar{2}gR}{2}​→v^{2}=\frac{2gR​}{4}\)
Or \(v^{2}=\frac{gR}{2}​\)
Now, putting the value of \(v^{2}\) in Eq. (i), we get
\(h=\frac{​\frac{gR}{2}​R​}{2gR−\frac{gR}{2}}\)
\(=\frac{\frac{gR^{2}}{R}}{\frac{3gR}{2}}\)​
Or \(h=\frac{R}{3}\)

Answer 3

The correct option is (B): \(\frac{R}{3}\)
Let H be the maximum height.
By conservation of energy,
\(T.E_{i}=T.E_{f}\)
\(K.E_{i}+P.E_{i}=P.E_{f}\)
\(\frac{1}{2}m\frac{Ve^{2}}{4}+(\frac{−GMm}{R})=\frac{−GMm}{R+H}\)
\(\frac{m2GM}{8R}−\frac{GMm}{R}=−\frac{GMm}{R+H}\) \((∵V_{e^{2}}=\frac{2GM}{R})\)
\(\frac{−3GMm}{4R}=\frac{−GMm}{R+H}\)
\(R+H=\frac{4R}{3}\)
\(∴H=\frac{R}{3}\)