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VITEEE 2009 Paper

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Question : 7 of 120

Marks: +1, -0

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

$\frac{R}{6}$
$\frac{R}{3}$
$\frac{2R}{3}$
R

Solution:

Here, maximum height attained by a projectile

h=\frac{v^{2}R}{2gR-v^{2}}

...(i)

Velocity of body = half the escape velocity

i.e.,

\;v=\frac{v_{e}}{2}

or

\;\;\;v=\frac{\sqrt{2gR}}{2}

⇒

\;\;\;v^{2}=\frac{2gR}{4}

⇒

\;\;\;v^{2}=\left[\frac{gR}{2}\right]

Now, putting value of

v^{2}

in Eq.(i), we get

Height,

\;\;h=\frac{\frac{gR}{2}\cdot R}{2gR-\frac{gR}{2}}=\frac{\frac{gR^{2}}{2}}{\frac{3gR}{2}}=\frac{R}{3}

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