Question

A projectile is projected with velocity $k v _ { e }$ in vertically upward direction from the ground into the space. (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)

• A. $\frac { R } { k ^ { 2 } + 1 }$
• B. $\frac { R } { k ^ { 2 } - 1 }$
• C. $\frac { R } { 1 - k ^ { 2 } }$
• D. $\frac { R } { k + 1 }$

Answer 1

Answer: (C)

$\frac { R } { 1 - k ^ { 2 } }$

Answer 2

From the law of conservation of energy

Difference in potential energy between ground and maximum height = Kinetic energy at the point of projection

$\frac { m g h } { 1 + h / R } = \frac { 1 } { 2 } m \left( k v _ { e } \right) ^ { 2 }$ $= \frac { 1 } { 2 } m k ^ { 2 } ( \sqrt { 2 g } R ) ^ { 2 }$

[As ]

By solving height from the surface of earth $h = \frac { R k ^ { 2 } } { 1 - k ^ { 2 } }$

So height from the centre of earth $r = R + h = R + \frac { R k ^ { 2 } } { 1 - k ^ { 2 } }$

$= \frac { R } { 1 - k ^ { 2 } }$.