Question

A particle of mass $'m'$ is projected with a velocity $v= kV _{ e }( k <1)$ from the surface of the earth.
$\left( V _{ e }=\right.$ escape velocity $)$
The maximum height above the surface reached by the particle is:

• A. $R\left(\frac{k}{1-k}\right)^{2}$
• B. $R \left(\frac{ k }{1+ k }\right)^{2}$
• C. $\frac{ R ^{2} k }{1+ k }$
• D. $\frac{ Rk ^{2}}{1- k ^{2}}$

Answer 1

Answer: (D)

As the particle is projected with a velocity less than escape velocity, it will go to a maximum height and come back.
From conservation of energy
$\frac{m g R h}{R+h}=\frac{1}{2} m\left(k V_{e}\right)^{2}$
$\frac{2 g R h}{R+h}=k^{2}(2 g R)$
$\frac{h}{R+h}=k^{2}$
$\frac{R+h}{h}=\frac{1}{k^{2}}$
$\frac{R}{h}+1=\frac{1}{k^{2}}$
$\frac{R}{h}=\frac{1}{k^{2}}-1$
$h=\frac{R k^{2}}{1-k^{2}}$