Q: A body starts from rest and falls vertically from a height of 19.6 metres. If g = 9.8 m/sec^{2}, 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

Ans:(b)

Sol: u = 0,a = 9.8 m/s^{2},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 $

t^{2} = 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