Q: A body starts from rest and falls vertically from a height of 19.6 metres. If g = 9.8 m/sec2, then the distance travelled by the body in the last 0.1 second of its motion is
(a) 0.049 m
(b) 1.9 m
(c) 17.7 cm
(d) 19.6 m
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Ans: (b)
Sol: u = 0,a = 9.8 m/s2,S = 19.6 m
$\displaystyle S = u t + \frac{1}{2}a t^2 $
$ \displaystyle 19.6 = 0 + \frac{1}{2}(9.8) t^2 $
t2 = 19.6/4.9 = 4
t = 2 sec
Let us calculate distance travelled in 1.9 s
$ \displaystyle S = u t + \frac{1}{2}a t^2 $
$\displaystyle S’ = 0 + \frac{1}{2}(9.8) (1.9)^2 $
S’ = 17.69 m
Distance travelled in last 0.1 s
= S- S’ = 19.6-17.69
=1.91 m