Physics - Gravitation Question with Solution | TestHub

The minimum velocity of projection to achieve escape velocity can be calculated as,

Initial$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$

$=\frac{1}{2}\times \mathrm{m}{\left(4\times 11.2\right)}^2=16\times \frac{1}{2}{\mathrm{mv}}_{\mathrm{e}}^2$

As$\frac{1}{2}{\mathrm{mv}}_{\mathrm{e}}^2\mathrm{ }$energy is used up in coming out from the gravitational pull of the earth, so

Final$\mathrm{ }\mathrm{KE}\mathrm{ }$should be$\mathrm{ }15\times \frac{1}{2}{\mathrm{mv}}_{\mathrm{e}}^2$

Hence,$\frac{1}{2}\mathrm{m}{\left({\mathrm{v}}^{\text{'}}\right)}^2=15\times \frac{1}{2}{\mathrm{mv}}_{\mathrm{e}}^2$

$\therefore {\mathrm{v}}^{\mathrm{\prime }2}=15{\mathrm{v}}_{\mathrm{e}}^2$

$\mathrm{or} {\mathrm{v}}^{\mathrm{\prime }}=\sqrt{15}{\mathrm{v}}_{\mathrm{e}}$

$=\sqrt{15}\times 11.2 \mathrm{km} {\mathrm{s}}^{-1}$