Q. A body starts from rest and falls vertically from a height of 19.6m. If $g = 9.8{ms}^{- 2}$, then the distance travelled by the body in the last 0.1s of its motion is
A) 1.9 m
B) 0.049 m
C) 17.7 m
D) 19.6 m
Explanation
(a) $t = \sqrt{\frac{2 \times 19.6}{9.8}} = 2s\ ;1.9 = \sqrt{\frac{2 \times h}{9.8}}$
or $h = \frac{1.9 \times 1.9 \times 9.8}{2}$m=17.689m
Required distance $= 19.6$ $- 17.689$ m = 1.9m.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.