Question

A body is projected upwards with a velocity of $4 \times 11.2\, km\, s^{-1}$ from the surface of earth. What will be the velocity of the body when it escapes from the gravitational pull of earth?

• A. $11.2\, km\, s^{-1}$
• B. $2 \times 11.2\, km\, s^{-1}$
• C. $3 \times 11.2\, km\, s^{-1}$
• D. $\sqrt{15} \times 11.2\, km\, s^{-1}$

Answer 1

Answer: (D)

lnitiaI KE of the body $= \frac{1}{2} mv^2$
Here, $ v = 4 \times 11.2\, kms^{-1}$
So, $ KE = \frac{1}{2} \times m (4 \times 11.2)^2$
$ = 16 \times \frac{1}{2} mv^2_e$
As $\frac{1}{2} mv^2_e$ energy is used up in coming out from the gravitational pull of the earth, so final KE should be
$15 \times \frac{1}{2} mv^2_e$
Hence, $ \frac{1}{2} mv' ^2 = 15 \times \frac{1}{2} mv^2_e$
$\therefore v'^2 = 15 v^2_e $
or $ v = \sqrt{15}v_e = \sqrt{15} \times 11.2\, kms^{-1}$