Q. A body is projected with velocity of $2 \times 11.2\ km/s$ from the form the surface of earth. The velocity of the body when it escapes the gravitational pull of earth is
A) $\sqrt{3} \times 11.2\ km/s$
B) $11.2\ km/s$
C) $\sqrt{2} \times 11.2\ km/s$
D) $0.5 \times 11.2\ km/s$
Explanation
(a) $KE = \frac{1}{2}{mv}^{2} - \frac{1}{2}m(11.2)^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}m(2 \times 11.2)^{2} - \frac{1}{2}m(11.2)^{2}$
$\frac{1}{2}{mv}^{2} = 3 \times \frac{1}{2}m \times (11.2)^{2}$
$\ \ \ \ \ \ \ \ \ v = \sqrt{3}\ \times 11.2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.