A body is projected vertically up with a velocity equal to $\frac{3}{4}$ of the escape velocity from the surface of the earth. The maximum height it reaches is: (Radius of the earth = R) [Height is measured from surface of earth]. If your answer is $\frac{aR}{b}$ (Report the value of a + b (a and b are coprime positive integers) 

Escape velocity ${\upsilon }_e\sqrt{\frac{2GM}{R}}.$

Velocity of projection $\upsilon =\frac{3}{4}{\upsilon }_e=\frac{3}{4}\sqrt{\frac{2GM}{R}}.$

The total energy of the body when it is projected is, 

$E_i=KE+PE=\frac{1}{2}m{\upsilon }^2-\frac{GmM}{R}=\frac{1}{2}m\times \frac{9}{16}\times \frac{2GM}{R}-\frac{GmM}{R}=\frac{9}{16}\frac{GmM}{R}-\frac{GmM}{R}=-\frac{7}{16}\frac{GMm}{R}$

Let h be the maximum height attained by the body.

The distance of the body from the centre of the earth is $r=R+h.$ At this height, the total energy of the body is, $E_f=KE+PE=0-\frac{GmM}{r}=-\frac{GmM}{r}=-\frac{GmM}{(R+h)}$

From the principle of conservation of energy, $E_i=E_f,$ i.e. $-\frac{7}{16}\frac{GmM}{R}=-\frac{GmM}{(R+h)}$

Or $7(R+h)=16Ror7h=9R$ or $h=\frac{9R}{7}$