Question

please someone help me in solving this question?

Answer 1

Answer: Option (C) is CORRECT.

Explanation:

Kinetic energy given to rocket at the surface of earth = Change in potential energy of the rocket in reaching from ground to highest point.

Solution is in the attachment.

Hope it helps ^_^

Answer 2

Given:

Rocket of mass M is launched from the surface of the earth with initial speed V. The radius of Earth is R and air resistance is negligible.

To find:

Maximum height attained by rocket

Concept:

The basic method to solve this type of questions is to to apply conservation of mechanical energy. The total mechanical energy of the rocket on the surface of the earth shall be equal to the total mechanical energy of the rocket at the maximum height.

Calculation:

Let mass of earth be m and max height reached be h .

Applying Conservation of Mechanical Energy :

$\therefore \: - \dfrac{GMm}{R} + \dfrac{1}{2} M {V}^{2} = - \dfrac{GMm}{(R + h)} + 0$

$= > \dfrac{1}{2} M {V}^{2} = GMm \bigg \{ \dfrac{1}{R} - \dfrac{1}{R + h} \bigg \}$

$= > \dfrac{1}{2} M {V}^{2} = GMm \bigg \{ \dfrac{R + h -R }{R(R + h)} \bigg \}$

$= > \dfrac{1}{2} \cancel{M} {V}^{2} = G \cancel{M}m \bigg \{ \dfrac{h }{R(R + h)} \bigg \}$

$= > \dfrac{1}{2} {V}^{2} = G m \bigg \{ \dfrac{h }{ {R}^{2} + h R} \bigg \}$

$= > \dfrac{ {R}^{2} + hR}{h} = \dfrac{2Gm}{ {V}^{2} }$

$= > \dfrac{ {R}^{2} }{h} +R = \dfrac{2Gm}{ {V}^{2} }$

$= > \dfrac{ {R}^{2} }{h} = \dfrac{2Gm}{ {V}^{2} } - R$

$= > h = \dfrac{ {R}^{2} }{ \bigg( \dfrac{2Gm}{ {V}^{2} } -R \bigg) }$

$= > h = \dfrac{ R }{ \bigg( \dfrac{2GmR}{ {R}^{2} {V}^{2} } -1 \bigg) }$

$= > h = \dfrac{ R }{ \bigg \{ \bigg(\dfrac{Gm}{ {R}^{2} } \bigg) \dfrac{2R}{ {V}^{2} } -1 \bigg \} }$

$= > h = \dfrac{ R }{ \bigg( \dfrac{2gR}{ {V}^{2} } -1 \bigg) }$

So , final answer is :

$\boxed{ \red{ \bold{ h = \dfrac{ R }{ \bigg( \dfrac{2gR}{ {V}^{2} } -1 \bigg) } }}}$